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OSMANIA UNIVERSITY LIBRARY 

Call No. S 7 *' &/DS$> Accession No. 



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last marked below. 



SPHERICAL TRIGONOMETRY 

AFTER THE CESARO METHOD 




SPHERICAL 
TRIGONOMETRY 

AFTER THE CESARO METHOD 




J. D. H. DONNAY. E.M., Ph.D. 



1945 

INTERSCIENCE PUBLISHERS. INC. - NEW YORK. N. Y. 



Copyright, 1945, by 

INTERSCIENCE PUBLISHERS, INC. 
215 Fourth Avenue, New York 3, N. Y. 



Printed in the United State* of America 
by the Lancaster Press, Lancaster, Pa. 



To THE MEMORY OF 

GIUSEPPE CESARO 

1849 - 1939 

CRT8TALLOGRAPHER AND MINERALOGIST 
PROFESSOR AT THE UNIVERSITY OF LIEGE 



PREFACE 

From a practical standpoint spherical trigonometry is 
useful to engineers and geologists, who have to deal with 
surveying, geodesy, and astronomy; to physicists, chemists, 
mineralogists, and metallurgists, in their common study of 
crystallography; to Navy and Aviation officers, in the solu- 
tion of navigation problems. For some reason, however, 
spherical trigonometry is not recognized as a regular subject 
in many American college curricula. As a consequence, the 
teacher of a science for which a working knowledge of spher- 
ical triangles is desirable usually finds he has to impart it 
to the students himself, or to use ready-made formulae that 
his listeners have never seen, let alone derived, before. 

This book has been written in an attempt to meet this 
situation. It can be covered in about ten to twelve lectures 
and could well serve as a text in a one-unit course for a 
quarter or a semester. It aims at giving the strict minimum, 
as briefly as possible. The straightforward and time-saving 
Cesaro method seems particularly suitable for this purpose, 
tying together as it does, from the outset, the concepts of 
spherical and plane trigonometry. This method has with- 
stood the test of experience. For years Belgian students 
have thrived on it. Personally I have taught it for eight 
years to crystallography students at the Johns Hopkins 
University, and for two years to freshmen classes at Laval 
University, with gratifying results. 

The order in which the subject matter is arranged may 
appear unorthodox. It has proved satisfactory, however, 
from the teaching point of view. Through the use of the 
stereographic projection (Ch. I), the concept of spherical 
excess somewhat of a stumbling block is mastered from 

vii 



viii PREFACE 

the start. A working knowledge of Cesaro's key-triangles is 
acquired, as soon as they are established (Ch. II), through 
the derivation of Napier's and Delambre's formulae and 
several expressions of the spherical excess (Ch. III). This 
much insures the understanding of the method. The treat- 
ment of the oblique-angled triangle (Ch. IV) is thereby so 
simplified that one can dispense with that of the right-angled 
triangle. The latter properly follows as a particular case 
(Ch. V). Examples of computations are given (Ch. VI), 
both with logarithms and with the calculating machine. 
They are followed by a selection of problems, completely 
worked out (Ch. VII), and a number of exercises with answers 
(Ch. VIII). Cross-references are made in the theoretical 
chapters to appropriate applications. Proofs believed to be 
new are marked by asterisks. 

In conclusion may I be permitted to say that this booklet 
was planned jointly by Cesaro and myself, several years ago. 
He who was to have been the senior author passed away 
shortly afterwards. In introducing Cesaro's elegant method 
to the English-speaking public, I would like to think that 
my writing will reflect his influence and, to some degree at 
least, his reverence for simplicity and rigor. 

J. D. H. D. 



Hercules Powder Company Experiment Station 

Wilmington, Delaware 

April 1945 



CONTENTS 

PAGE 

Dedication v 



Preface . 



vn 



Introduction 3 

1. Purpose of Spherical Trigonometry 3 

2. The Spherical Triangle 4 

3. Comparison between Plane and Spherical Trigo- 
nometry 5 

4. Fields of Usefulness of Spherical Trigonometry. 5 

I. The Stereographic Projection 7 

1. Definition 7 

2. First Property of the Stereographic Projection. 7 

3. Second Property of the Stereographic Projection. 10 

II. Cesaro's Key-Triangles 13 

1. Cesaro's "Triangle of Elements" 13 

2. Cesaro's "Derived Triangle" 16 

3. Polar Triangles 18 

4. Derivation of the Formulae of Spherical Trigo- 
nometry 20 

Exercises 21 

III. How the Key-Triangles are Put to Work 22 

1. Napier's Analogies 22 

2. Delambre's Formulae 22 

3. Euler's Formula 23 

4. Lhuilier's Formula 24 

5. Cagnoli's Formula 25 

iz 



CONTENTS 

PAGE 

6. The Spherical Excess in Terms of Two Sides and 

Their Included Angle 26 

Exercises 26 

IV. Relations between Four Parts of a Spherical Tri- 
angle 28 

1. Relation between the Three Sides and One Angle 28 

2. Expression of the Half-Angles in Terms of the 
Three Sides 29 

3. Relation between Two Sides and Their Opposite 
Angles 29 

4. Relation between the Three Angles and One Side 30 

5. Expression of the Half-Sides in Terms of the 
Three Angles 30 

6. Relation between Two Sides, their Included 
Angle, and the Angle Opposite One of them .... 32 

Exercises 34 

V. Right-Angled Triangles 36 

1. The Right-Angled Triangle, a Special Case of 
the Oblique-Angled Triangle 36 

2. Relations between the Hypotenuse and Two 
Sides or Two Angles (the product formulae) .... 37 

3. Relations between the Hypotenuse, One Side, 
and One Angle (the ratios of sines and tangents) . 37 

4. Relations between Two Angles and One Side, or 
between Two Sides and One Angle (the ratios of 
cosines and cotangents) 38 

5. Napier's Rule 39 

VI. Examples of Calculations 41 

1. Solution of Triangles 41 

2. Formulae Adapted to Logarithmic Computation 42 

3. Numerical Calculations 44 

4. First Example: given a, 6, c, solve for A 45 

5. Second Example: given A, B, C, solve for a, 6, c. 48 



CONTENTS zi 

PAGE 

6. Third Example: given a, 6, and C = 90, solve 
for A 50 

7. Fourth Example: spherical distance between two 
points given by their geographical coordinates. . 51 

8. Numerical Application 52 

VII. Problems (Selected Problems With Complete Solu- 
tions) 55 

VIII. Exercises 70 

Answers to Exercises 74 

Appendix 1. Spherical Areas 76 

Appendix 2. Formulae of Plane Trigonometry 78 

Index 81 



SPHERICAL TRIGONOMETRY 

AFTER THE CESARO METHOD 



INTRODUCTION 

1. Purpose of Spherical Trigonometry. Spherical trigonome- 
try is essentially concerned with the study of angular rela- 
tionships that exist, in space, between planes and straight 
lines intersecting in a common point O. A bundle of planes 
passed through intersect one another in a sheaf of straight 
lines. Two kinds of angles need therefore be considered: 
angles between lines l and angles between planes (dihedral 
angles). 

The spatial angular relationships are more easily visualized 
on a sphere drawn around O with an arbitrary radius. Any 
line through is a diameter, and any plane through a 
diametral plane, of such a sphere. The former punctures the 
sphere in two diametrically opposite points, the latter inter- 
sects it along a great circle. The angle between two lines is 
measured on the sphere by an arc of the great circle whose 
plane is that of the two given lines. The angle between two 
planes is represented by the angle between the two great 
circles along which the given planes intersect the sphere. 
Indeed, by definition, the angle between the great circles is 
equal to the angle between the tangents to the circles at their 
point of intersection, but these tangents are both perpen- 
dicular to the line of intersection of the two given planes, 
hence the angle between the tangents is the true dihedral 
angle. 

An open pyramid, that is to say a pyramid without base, 
whose apex is made the center of a sphere determines a 
spherical polygon on the sphere. The vertices of the polygon 
are the points where the edges of the pyramid puncture the 

1 From now on, the word line will be used to designate a straight 
line, unless otherwise stated. 



4 INTRODUCTION 

sphere; the sides of the polygon are arcs of the great circles 
along which the faces of the pyramid intersect the sphere. 
The angles of the polygon are equal to the dihedral angles 
between adjacent faces of the pyramid. The sides of the 
polygon are arcs that measure the angles of the faces at the 
apex of the pyramid, that is to say, angles between adjacent 
edges. 

A trihedron is an open pyramid with three faces. The 
three axes of co-ordinates in solid analytical geometry, for 
instance, are the edges of a trihedron, while the three axial 
planes are its faces. Consider a trihedron with its apex at 
the center of the sphere. It determines a spherical triangle 
on the sphere. 2 In the general case of an oblique trihedron, 
an oblique-angled spherical triangle is obtained, that is to 
say, one in which neither any angle nor any side is equal to 
90. The main object of spherical trigonometry is to investi- 
gate the relations between the six parts of the spherical tri- 
angle, namely its three sides and its three angles. 

2. The Spherical Triangle. The sides of a spherical triangle 
are arcs of great circles. They can be expressed in angular 
units, radians or degrees, since all great circles have the same 
radius, equal to that of the sphere. As a consequence of the 
conventional construction by means of which the spherical 
triangle has been defined (Sn. 1), any side must be smaller 
than a semi-circle and, likewise, any angle must be less 
than 180. 

By considering the trihedron whose apex is at the center of 
the sphere, we see that the sum of any two sides of a spherical 
triangle is greater than the third side, that any side is greater 
than the difference between the other two sides, and that the 
sum of all three sides (called the perimeter) is less than 360. 

Because any angle of a spherical triangle is less than 180, 
the sum of all three angles is obviously less than 540. It is 
greater than 180, as we shall see later (Ch. II, Sn. 1). 

This is the "Eulerian" spherical triangle, the only one to be 
considered in this book. 



USEFULNESS OF SPHERICAL TRIGONOMETRY 5 

3. Comparison between Plane and Spherical Trigonometry. 

In plane trigonometry, you draw triangles in a plane. Their 
sides are segments of straight lines. The shortest path from 
one point to another is the straight line that connects them. 
The distance between two points is measured, along the 
straight line, in units of length. The sum of the angles of a 
plane triangle is 180. Through any point in the plane, a 
straight line can be drawn parallel to a given line in the plane 
(Euclidian geometry). 

In spherical trigonometry, triangles are drawn on a sphere. 
Their sides are arcs of great circles. The shortest path (on 
the sphere) from one point to another is the great circle that 
connects them. The (spherical, or angular) distance between 
two points is measured, along the great circle, in units of 
angle. The sum of the angles of a spherical triangle is greater 
than 180. Through a point on the sphere, no great circle 
can be drawn parallel to a given great circle on the sphere 
(spherical geometry is non-Euclidian). 

4. Fields of Usefulness of Spherical Trigonometry. Most 
problems dealing with solid angles can be reduced to questions 
of spherical trigonometry. Such problems crop up in the 
study of geometrical polyhedra. Problems of solid analytical 
geometry in which planes and lines pass through the origin 
usually have trigonometric solutions. 

Problems involving spatial directions around one point are 
encountered in crystallography. Well formed crystals are 
bounded by plane faces. From a point taken anywhere 
inside the crystal, drop a perpendicular on each face; this 
face normal defines the direction of the face. Relationships 
between the inclinations of the faces relative to one another 
appear in the network of spherical triangles which the sheaf 
of face normals determines on a sphere drawn around 0. 

Surveying is concerned with such small regions of the earth 
surface that they can be considered plane in a first approxima- 
tion. Geodesy deals with larger regions, for which the curva- 



6 INTRODUCTION 

ture of the earth must be taken into account. In a second 
approximation the earth is taken as spherical, and the for- 
mulae of spherical trigonometry are applicable. (Further re- 
finements introduce corrections for the lack of perfect sphe- 
ricity of the "geoid.") 

In astronomy the application of spherical trigonometry is 
obvious. The observer occupies a point that is very nearly 
the center of the celestial sphere around the earth. Each 
line of sight is a radius of the sphere. To the observer who is 
not aware of, or concerned with, the distances from the earth 
to the heavenly bodies, the latter appear to move on a sphere. 
The angle subtended by two stars, as seen by the observer, 
will thus become a side in a spherical triangle. Navigation 
techniques, either on the high seas or in the air, being based 
on astronomical observations, likewise depend on the solution 
of spherical triangles. 



CHAPTER I 

THE STEREOGRAPHIC PROJECTION 

1. Definition. The problem of representing a sphere on a 
plane is essentially that of map projections. Of the many 
types of projections that have been devised, one of the most 
ancient is the stereographic. 

In geographical parlance, used for convenience, the pro- 
jection plane is the plane of the equator, and the projection 
point, the South Pole. Points in the Northern Hemisphere 
are projected inside the equatorial circle; points in the South- 
ern Hemisphere, outside the equatorial circle; any point on 
the equator is itself its own stereographic projection (Fig. 1). 
The North Pole is projected in the center of the projection; 
the South Pole, at infinity. 




FIG. 1. The stereographic projection. 



2. First Property of the Stereographic Projection. Circles 
are projected as circles or straight lines. If the circle to be 
projected passes through S, its projection is a straight line. 
This is obvious, since the projection of the circle is the inter- 

7 



8 I. THE STEREOGRAPHIC PROJECTION 

section of two planes: the plane of the circle and the plane of 
the equator. Note that if the given circle is a great circle 
passing through S (hence, a meridian) its projection is a 
diameter of the equator. ' 

If the circle to be projected does not pass through S, its 
projection is a circle. The proof 1 of this is based on the 
following theorem. 

Consider (Fig. 2) an oblique cone with vertex S and circular 
base AB. Let the plane of the drawing be a section through 
S and a diameter AB of the base. The circular base is pro- 
jected on the drawing as a straight line AB. A section ab, 
perpendicular to the plane of the drawing, and such that the 
angles SAB, S6a are equal, is called sub-contrary (or anti- 
parallel) to the base. 

Theorem: In an oblique cone having a circular base, any 
section sub-contrary to the base is circular. 




FIG. 2. Sub-contrary sections. 



Take a section cd parallel to the base and, hence, obviously 
circular. The two sections ab and cd intersect along a com- 
mon chord, projected at a point n, which bisects the chord. 
Let x be the length of the semi-chord. The triangles can and 
bdn are similar (angles equal each to each), hence an : nd 

1 This proof can be omitted in a first reading. 



FIRST PROPERTY 9 

= cn : nb, or an.nb = cn.nd. Because cd is circular, cn.nd 
= x 2 . Hence an . nb = x 2 , and ab is also circular. 2 

Now consider a section of the sphere of projection cut 
perpendicular to the intersection of the equatorial plane EE' 
and the plane of the circle AB to be projected (Fig. 3). 
The angles SAB, S6a (marked on the drawing) are equal, 
since the measure of SAB = KSE' + BE 7 ) and that of 
S6a = |(SE + BE') are equal (because SE = SE' = 90). 
The sections ab and AB of the cone of projection are therefore 
sub-contrary. Since AB is circular, so is ab. 




S 
FIG. 3. Projected circle, a circle. 

Remark. The center of the projected circle is the projec- 
tion of the vertex of the right cone tangent to the sphere along 
the given circle. Let C be the vertex of the right cone tangent 
to the sphere along the given circle AB (Fig. 4). Join CS, 
intersecting the equator in c and the sphere in D. The 
angles marked a are equal as having the same measure (one 
half arc AD). Likewise for the angles marked 0, 7, d. The 
Law of Sines, applied to the triangles Sac and Sc6, gives: 

ac : Sc = sin a : sin 0, cb : Sc = sin 7 : sin 5. 

2 This reasoning rests on the theorem, "If, from any point in the 
circumference, a perpendicular is dropped on a diameter of a circle, 
the perpendicular is the mean proportional of the segments deter- 
mined on the diameter," and its converse. 



10 I. THE STEREOGRAPHIC PROJECTION 

Applied to the triangles ACD and BCD, the same law gives: 
sin a : sin ft = CD : CA, sin 7 : sin 5 = CD : CB. 

Since CA = CB (tangents drawn to the sphere from the same 
point), these ratios of sines are equal. Whence ac = c6, and 
c is the center of the projected circle. 3 




FIG. 4. Center of projected circle. 

Note that if the given circle is a great circle (but not a 
meridian, nor the equator itself) its projection will be a circle 
having a radius larger than that of the equator and cutting 
the equator at the ends of a diameter. A great circle cuts the 
equator at the ends of a diameter of the latter; points on the 
equator are themselves their own stereographic projections. 

3. Second Property of the Stereographic Projection. The 

angle between two circles is projected in true magnitude.* The 

8 The proof holds good if the given circle is a great circle; the right 
cone with circular base becomes a right cylinder with circular base. 
Make the construction. 

4 A more general property of the stereographic projection is that 
the angle between any two curves on the sphere is projected in true 
magnitude. The property proved in the text, however, is sufficient 
for our purpose. This proof can be omitted in a first reading. 



SECOND PROPERTY 



11 



angle between two circles drawn on a sphere is equal to the 
angle between their tangents. We shall prove: (1) that the 
angle between the projected tangents is equal to the angle 
between the tangents; (2) that the projected tangents are 
tangent to the projected circles. Thus will be established 
the property that the angle between the projected circles is 
equal to the angle between the circles. 




Fia. 5. Angle between projected tangents equal to 
angle between tangents. 

(1) Consider (Fig. 5) the point P in which two given circles 
intersect. Let PT and PT' be the tangents to these circles; 
they cut the plane of the equator in <, t' and the plane tangent 
to the sphere at S in T, T'. Join PS, intersecting the plane 
of the equator in p y the stereographic pole of P. The pro. 
jected tangents are pi, pt f . Join ST and ST'. 

The triangles TPT' and TST' are similar (TT common; 
TP = TS and TT = T'S, as tangents drawn to the sphere 
from the same point). Hence, angle TPT' = angle TST'. 
Again, the triangles tpt' and TST' are similar (all sides parallel 
each to each; two parallel planes being intersected by any 



12 I. THE STEREOGRAPHIC PROJECTION 

third plane along parallel lines). Hence, angle tpt' = angle 
TST'. It follows that the angle between the tangents (TPT') 
and the angle between the projected tangents (tpt') are equal. 




s 
FIG. 6. Projected tangent, tangent to projected circle. 

*(2) Consider (Fig. 6) a right cone tangent to the sphere 
along the given circle PBC; let A be the vertex of this cone. 
We know that the circle PBC is projected as a circle pbc. 
We have seen that the projection of the vertex A is the center 
a of the projected circle. A generatrix AP of the right cone 
is projected as a radius ap of the projected circle. Pt, the 
tangent to the given circle at P, is projected in pt (t in the 
plane of the equator). But the angle apt, being the angle 
between the projections of the tangents PA and Pt, is equal 
to the angle between the tangents themselves, that is to say 
90. Hence, pt is tangent to the projected circle. 

* Proofs believed to be new are marked by asterisks. 



CHAPTER II 

CESARO'S KEY-TRIANGLES 1 

1. Ces&ro's "Triangle of Elements." Consider a spherical 
triangle ABC. Without loss of generality, we may suppose 
that one of its vertices A is located at the North pole of the 
sphere. Project this triangle stereographically. The pro- 
jected triangle A'B'C' (Fig. 7) will have a vertex A' at the 




FIG. 7. Stereographic projection. 

center of the projection, and two of its sides, A'B' and A'C', 
being projected meridians, will be straight lines; its third side 
B'C' will be an arc of a circle. 

At B' and C' draw the tangents to the circle B'C', meeting 
in T. Because the stereographic projection is angle-true, 

C'A'B' = A, A'B'T = B, A'C'T = C. 

1 Ces&ro, G. Nouvelle mthode pour l^tablissement des formules 
de la trigonometric sphe*rique. Bull. Acad. roy. de Belgique (Cl. des 
Sc.), 1905, 434. 

Les formules de la trigonometric sphe"rique de"duites de la 

projection stereographique du triangle. Emploi de cette projection 
dans les recherches sur la sphere. Bull. Acad. roy. de Belgique (Cl. 
des Sc.), 1905, 560. 

13 



14 



II. CESARO'S KEY-TRIANGLES 



Designate by 2E the external angle between the tangents 
meeting in T. It is easy to see that 



A + B + C - 180 



2E. 



The angle 2E is called the spherical excess of the spherical 
triangle ABC. It is equal to the excess over 180 of the sum 
of the angles of the spherical triangle. We shall express it 
in degrees. 

The angles of the plane triangle A'B'C' are expressed as 
follows, in terms of the angles of the spherical triangle ABC 
and its spherical excess 2E, 

A' = A, B' = B - E, C' = C - E. 

The sides of A'B'C' are functions of the sides of the spher- 
ical triangle ABC (Fig. 8). Taking the radius of the sphere 




FIG. 8. Perspective drawing of the 
sphere of projection . 



as the unit of length, we have 



c' = tan - , b' = tan - . 



"TRIANGLE OF ELEMENTS" 15 

Each of the quadrilaterals ABB'A' and ACC'A' has two 
opposite angles equal to 90 (one at A', by construction; the 
opposite one, as being inscribed in a semi-circle) and is, 
therefore, inscribable in a circle. Hence 

SB. SB' = SA.SA' = SC.SC' 
and BCC'B' are also concyclic. It follows that the angles 




.In I 
Fio. 9. Triangle of elements relative to the angle A. 

marked /3 (and the angles marked 7) are equal, so that the 
triangles SBC and SC'B' are similar. 
We may write, therefore, 

B'C' : BC = SC' : SB, 

or 2 

b 

, sec H 

a 2 



2 sin jr 2 cos 



whence 



a= 



c 
cos cos 



8 For those who prefer step-by-step derivations: B'C' a', by 
definition. BC = 2 sin Ja, for the chord BC subtends an arc a and 
the chord is equal to twice the sine of half the angle. SC', in the 
right-angled triangle SC'A', where SA' 1, is the secant of the angle 
A ? SC'. Finally, in the triangle SAB, where the angle at B is a right 
angle, SB - SA cos ASB - 2 cos Jc. 



16 II. CESARO'S KEY-TRIANGLES 

The triangle obtained by multiplying the three sides of 
A'B'C' by 

b c 
cos cos ^ 

is Cesaro's triangle of elements relative to the angle A (Fig. 9). 
Other "triangles of elements/* relative to the angles B 
and C, can be obtained by cyclic permutations. 

2. Ces&ro's "Derived Triangle." A lune is the spherical sur- 
face bounded by two great semi-circles; for instance (Fig. 8), 




FIG. 10. The two complementary trihedra, each showing its 
six parts and spherical excess. 

the area ABSCA between two meridians, ABS and ACS. 
Two trihedra are called complementary when the two spherical 
triangles they determine on the sphere form a lune. For in- 
stance (Fig. 8), the trihedra A'ABC and A'SBC are comple- 
mentary. They have two edges, A'B and A'C, in common 
and the third edge, A'S, of one is the prolongation of the 
third edge, A A', of the other; the vertices A and S lie at the 
ends of a diameter, and the two spherical triangles ABC and 
SBC are seen to form a lune. 

Designating by O the center of the sphere, consider a 
trihedron OABC, represented (Fig. 10) by its spherical tri- 
angle ABC. Produce the great circles BA and BC till they 
meet, in D, thus forming a lune. The spherical triangle ADC 
represents the complementary trihedron OADC. The parts of 



'DERIVED TRIANGLE* 1 



17 



the triangle ADC are easily expressed in terms of those of 
the triangle ABC. One side, 6, is common; the angle at D 
is equal to B; the other parts are the supplements of corre- 
sponding parts of the triangle ABC (Fig. 10). The spherical 
excess is found to be (180 - A) + (180 - C) + B - 180 
= 180 - (A + B + C) + 2B = 2B - 2E = 2(B - E). 



cos 2 cos I 




FIG. 11. Derived triangle relative to the angle A. 

Let us compose the triangle of elements, relative to the 
angle (180 A), of the complementary trihedron OADC. 
Its six parts are tabulated below, together with those of the 
triangle of elements, relative to A, of the primitive trihedron 
OABC. 

THE SIX PARTS OF THE TRIANGLE OF ELEMENTS 



For trihedron OABC 



For trihedron OADC 



Angles 


Sides 


Angles 


Sides 


A 

BTT 


. a 


/1OAO A\ 


. 180 - a 


c 


. b c 


V.loU A) 

Bfft TT<\ 


. b 180 - 


CT7 1 


sin cos 7: 
. c b 


/1ono /~i\ /"R Tp\ 


. 180 - c 


b 
2 




sin cos ~ 







18 II. CESARO'S KEY-TRIANGLES 

The expression of the third angle is transformed as follows: 
180 -C-B+E=A+ 180 -(A + B + C) + E = A 
2E + E = A E. The other parts are easily simplified. 

The new key-triangle can now be drawn (Fig. 11); it is 
Cesar o's derived triangle relative to the angle A. The derived 
triangle of a trihedron is the triangle of elements of the 
complementary trihedron. 

Other "derived triangles/' relative to the angles B and C, 
can be obtained by cyclic permutations. 

3. Polar Triangles. Consider a spherical triangle ABC and 
the corresponding trihedron OABC. Erect OA*, OB*, OC*, 
perpendicular to the faces of the trihedron, OBC, OCA, OAB, 
and on the same sides of these faces as OA, OB, OC, respec- 
tively. The new trihedron OA*B*C* determines, on the 
sphere (Fig. 12), a triangle A*B*C*. The vertices, A*, B*, 
C*, are called the poles of the planes OBC, OCA, OAB, 
respectively. 




C* 

FIG. 12. Polar triangles. 

The triangle A*B*C* is said to be the polar triangle of ABC. 
It follows from the construction that ABC is the polar triangle 
of A*B*C*. Likewise, it follows that the angles of one tri- 
angle are the supplements of the sides of the other, and that 
the sides of one are the supplements of the angles of the other. 



POLAR TRIANGLES 19 

Hence the perimeter 2p* of A*B*C* is equal to 360 - 2E, 
and its spherical excess 2E* is equal to 360 2p, where 2E 
and 2p refer to the triangle ABC. In other words half the 
spherical excess of one triangle is the supplement of half the 
perimeter of the other. 

Both the triangle of elements and the derived triangle of the 
polar triangle A*B*C* are thus easily established (Figs. 13 




FIG. 13. Triangle of elements of the polar triangle (relative to a). 




FIG. 14. Derived triangle of the polar triangle (relative to a). 

and 14). Their sides and angles are functions of the parts of 
the triangle ABC, in particular of its semi-perimeter p. 

Analogous key-triangles, relative to b and c, can be obtained 
by cyclic permutations. 

Remark. It is easy to remember how to transform the 
key-triangles of the primitive triangle into those of the polar 
triangle. Instead of a function of a half-side, write the 



20 11. CESARO'S KEY-TRIANGLES 

a A 

co-function of the half-angle; thus sin ^ becomes cos ^ , 

Z 1 

sin x cos jr becomes cos ^ sin -^ , etc. Instead of an angle, 
z & & 

write the supplement of the side; thus A becomes (180 a), 
and (180 A) becomes a. Instead of half the spherical ex- 
cess, E, write the supplement of half the perimeter (180 p). 
Instead of an angle minus half the spherical excess, write half 
the perimeter minus the side; thus (A E) becomes (p a) 
etc. The latter transformation is immediately apparent, 
since A* - E* = (180 - a) - (180 - p). 

4. Derivation of the Formulae of Spherical Trigonometry. 

All the formulae of spherical trigonometry are derived from 
the key-triangles (which have been obtained without any 
restrictive hypothesis on the spherical triangle, and are there- 
fore perfectly general) by applying to them the known formu- 
lae of plane trigonometry. Each formula of spherical trigo- 
nometry can thus be derived independently of the others. 




FIG. 15. A plane triangle. 

A' + B' + C' = 180 

a' + b' + c' = 2p' 

The parts of a plane triangle (Fig. 15) will be designated 
by primed letters: A', B', C', the angles; a', 6', c', the opposite 
sides f 2p' = a' + V + c', the perimeter. 

The parts of a spherical triangle (Fig. 16) will be desig- 
nated by unprimed letters: A, B, C, the angles; a, 6, c, the 




21 



FIG. 16. A spherical triangle. 
A + B + C - 180 = 2E 
a + b + c = 2p 

sides; 2p = a + b + c, the perimeter; 2E = A + B + C 
180, the spherical excess. 

EXERCISES 

1. Derive the triangle of elements relative to the angle B from 
that relative to the angle A (Fig. 9) by cyclic permutations. 

2. Same question for the triangle of elements relative to C. 

3. Draw the derived triangles relative to the angles B and C. 

4. Derive, for the polar triangle, the triangle of elements and the 
derived triangle: (i) relative to the side b, (ii) relative to the side c. 

5. Show that the area of the triangle of elements and that of the 
derived triangle are both equal to A/8, where 

A =* 2Vsin p sin(p a) sin(p b) sin(p c). 
(A, called the sine of the trihedral angle, is a useful function of the 
face angles of the trihedron, which are the sides of the spherical 
triangle.) 



CHAPTER III 

HOW THE KEY-TRIANGLES ARE PUT TO WORK 

1. Napier's Analogies. 1 Napier's analogies are relations be- 
tween five parts of a spherical triangle. 



tan*(B-C) 


sin (& c) 


cot JA 


sin J(6+c)' 




tan %(b c) 


sin i(B-C) 


tan ia 


" sin J(B + C)' 



tan J(B-f C) 


cos i^c). 


cot JA cos Kb+c)' 




tan l(b-f-c) cos l(B-C) 


tan Ja 


cos i(B+Q* 



(1) 



(2) 



These formulae are read directly from the key-triangles, 
by applying the Law of Tangents: In a plane triangle, the 
tangent of the half-difference of two angles is to the tangent 
of their half-sum (or to the cotangent of half the third angle) 
as the difference of the opposite sides is to their sum. 

The triangle of elements (Fig. 9) yields the first analogy. 
Note that the half-sum of two angles is B C. We have 



. 

tan 



B C 



. b c . c b . b - c 
sin ^ cos ^ sin - cos ~ sm -^ 



.A 

cot 



.6 c , . c b 
sin ^ cos ^ -f sm ^ cos ~ 



sm 



b + c' 



The derived triangle (Fig. 11) gives the second analogy. The 
last two are obtained from the key-triangles (Figs. 1$ and 14) 
of the polar triangle. 

2. Delambre's Formulae. 2 Delambre's formulae are rela- 
tions between all six parts of a spherical triangle. 

1 Analogies is an archaic term for proportions. 

1 Improperly attributed to Gauss by certain authors. 



NAPIER'S AND DELAMBRE'S FORMULAE 



23 



co8j(B-C) ^ sin j(b+c) 
sin JA sin Ja ' 



cos J (B + C) ^ cos j(b-f c) 
sin JA cos Ja ' 



sin i(B-C) 


sin J(b c) > 


cos JA 


sin Ja ' 




sin i(B+C) 


cos J(6 c) 


cos JA 


cos Ja 



(3) 



(4) 



The Law of Sines of plane trigonometry, applied to the 
triangle of elements (Fig. 9), gives 



sm - 



. 6 c 
sin -cos 



. c b 
am cos 



sin A sin (B - E) sin (C - E) ' 
which, by the theory of proportions, becomes 



. b + c 

Sm :r 



sm 



b - c 



. a 

sm o 

2 _ z z 

sin A ~ sin(B-E) + sin(C-E) ~ sin(B-E) - sin(C-E) 
or 

a .b + c .... 6 c 



sm 2 



sm 



sm 



.A A A B-C 

2 Sm TT COS TT 2 COS 77 COS S 



. A . B - C' 

2 sm " sm 



and Delambre's first two formulae follow immediately. 

The same method, applied to the derived triangle (Fig. 11), 
yields the last two formulae. 

3. Euler's Formula. Euler's formula expresses E, one-half 
of the spherical excess, 3 in terms of the sides. 



cos E = 



cos a -f cos b + cos c 

. a b c 
4 cos cos ~ cos ~ 



(5a) 



3 The value E of one-half the spherical excess is useful in calculating 
the area of a triangle. It is known from geometry (see Appendix 1) 
that the area of a spherical triangle is to the area of the sphere as E 
(expressed in degrees) is to 360. 



24 III. KEY-TRIANGLES PUT TO WORK 

This formula is obtained from the derived triangle (Fig. 11), 
by applying the Law of Cosines of plane trigonometry 

a '2 = /2 + c > 2 _ ^V c' cos A'. 
We have 

. 6 . n c n d , 6 c rt a b c r. 

sm 2 sin 2 - = cos 2 ~ + cos 2 ~ cos 2 ~ 2 cos cos ~ cos ~ cos E, 

Z Z t & & & & & 

n a b c ^ 9 a 

2 cos cos ~ cos ~ cos E = cos 2 ~ 

JL u 

,/ 6 c . . 6 . c \/ 6 c . b . c\ 
+ ( cos ^ cos g + sm - sm - 1 ( cos ^ cos ^ - sm 3 sm 2 / ' 

A a b c -n o 9 a .o & C 6 + C 

4 cos H cos jc cos cos E = 2 cos 2 + 2 cos cos ^ , 

& & A & A 

from which Euler's formula follows immediately. 

4. Lhuilier's Formula. Lhuilier's formula gives the fourth of 
the spherical excess in terms of the sides (and the semi- 
perimeter p). 



x oE , p . p a , p b , p c 
tan 2 ^ = tan ^ tan ^r tan ^ ^ tan *-= 

L t & fj t 



(5b) 



It can be obtained directly from the derived triangle 
(Fig. 11) by expressing the tangent of half the angle E in 
terms of the sides. The plane trigonometry formula is 



- p'(p' - a') 
We have 

o f OL/ 6 c a 

2p 26 = cos ~ -- cos ~ , 

/u t 

r / 6 c , a 

2p' = cos -- + cos , 



FORMULAE FOR SPHERICAL EXCESS 25 

and 

r I r / a b + C 

2p 2c = cos K cos ^ . 



Since 



2p' 2a' = cos ^ + cos ^ , 



cos P - cos Q , P + Q , Q - P 
- - - = tan 1 r^- tan -* 



- - - 1 r- -= , 

cos P + cos Q 2 2 * 

we are led to 

, E . a + b c L a b + c 
tan 2 75- = tan - - - tan - - - 

. . . a + b 4- c , b + c a 
X tan - -r - tan - - - , 

which is Lhuilier's formula, as 2p = a + b + c. 

*5. Cagnoli's Formula. Cagnoli's formula gives the half of 
the spherical excess in terms of the sides (and the semi- 
perimeter p). 



. ^ Vsm p sm(p a) sin(p 6) sm(p c) 

sin Hi = r 

a b c 

2 COS jr COS COS 7 



(5c) 



The area T' of the derived triangle (Fig. 11) may be 
expressed as the half-product of the base by the altitude, 

T' = ? cos 5 cos ~ cos ~ sin E, 

u i t 

or, in terms of its sides and semi-perimeter, 4 



= i Vsin p sin(p a) sin(p b) sin(p c). 
4 Cp. Exercise 5, Chapter II and Section 4, this chapter. 



26 III. KEY-TRIANGLES PUT TO WORK 

Equating the two expressions immediately gives the desired 
formula. 

*6. The Spherical Excess in Terms of Two Sides and Their 
Included Angle. In a plane triangle, as a consequence of the 
Law of Sines, 

c' - V cos A' 



cot B' = 



b' sin A' 



This formula, applied to the derived triangle (Fig. 11), 
in which the angles (180 A), E, (A E) are taken as 
A', B', C', respectively, yields the desired relation 



, T-, 

cot E 



b c , . 6 . c . 
cos x cos JJT + sin ~ sin ^ cos A 

. 6 . c . A 
sin - sin jr sin A 

fL 



or 



cot - cot ~ + cos A 

cot E = 4^ 

sin A 



(6) 



EXERCISES 

1. Express cosp in terms of the angles, by applying Euler's 
formula to the polar triangle. 

2. Derive the expression of cos p in terms of the angles from the 
derived triangle of the polar triangle (Fig. 14), in the same way as 
Euler's formula has been derived (Sn. 3). 

3. From Napier's second analogy (1), derive a formula to calculate 
E in terms of two sides (6, c) and their included angle (A) . 

4. Show that, in a right-angled triangle (C = 90), 

tan E = tan - tan ^ . 



Hint: use the formula derived in the preceding exercise. 

E E 

5. Find sin ^ and cos jr- in terms of the sides. Apply the s 



method as that used in Sn. 4 for deriving Lhuilier's formula (6). 



EXERCISES 27 

6. Derive Lhuilier's formula (6) from the results obtained in the 
preceding exercise. 

7. Derive Cagnoli's formula from Lhuilier's formula. 

8. Gua's formula gives cot E in terms of the sides. Find what it is. 

9. You now know six formulae by means of which, given the sides, 
the spherical excess can be calculated. Which one would you prefer 
if you had to depend on logarithmic calculations? Which one would 
be easiest to use if a calculating machine were available? 

10. The sides of a spherical triangle are a = 3526', 6 = 4215', 
c 1822'. Calculate the spherical excess by two different formulae. 



CHAPTER IV 

RELATIONS BETWEEN FOUR PARTS 
OF A SPHERICAL TRIANGLE 

1. Relation between the Three Sides and One Angle. 1 Ap- 
plying the Law of Cosines of plane trigonometry 

a'* = fc'2 + c '2 _ ^V c' cos A' 
to the triangle of elements (Fig. 9), we get 

a b c c b 

sin 2 = sin 2 ^ cos 2 ~ + sin 2 ~ cos 2 ~ 



. b c . c b . 
2 sm cos H sin cos ~ cos A. 



In view of 



2 sin 2 - = 1 cos x, 2 cos 2 ~ = 1 + cos x\ 

t A 

the above formula, multiplied by 4, may be written 

2(1 cos a) = (1 cos b)(l + cos c) 

+ (1 cos c)(l + cos b) 2 sin b sin c cos A, 

whence the desired formula 

' (7) 



cos a cos b cos c + sin 6 sin c cos A. 



This is expressed: The cosine of the side opposite the given 
angle is equal to the product of the cosines of the other two 
sides, plus the product of the sines of these two sides times 
the cosine of the given angle. 

i Used in Ex. 16, 21, 24, 26, 28 (Ch. VIII). 



OBLIQUE-ANGLED TRIANGLES 



29 



2. Expression of the Half-Angles in Terms of the Three 
Sides. 2 The following formulae of plane trigonometry 



l 

2 



' - V)(p' - c') 
6V 



COS' 



A' _ p '(p> - fl ') 
2 6V 



p'(p> - a') 



are applied to the triangle of elements (Fig. 9). 
We have 

n , . b + c . . a n . p p a 
2p' = sm ^ -- h sin - = 2 sm | cos ^ , 



2p' 2a' = sin 



.a n . p a p , 
sm ^ = 2 sin - - cos ^ , etc. 



Hence, in the spherical triangle, 



. 2 A 
sm ^ 

2 A 
cos 2 2 - 

tan*^ 


sin(p - 6) sin(p - c) 


(8a) 
(8b) 
(8c) 


sin 6 sin c ' 
sin p sin(p a) 


sin 6 sin c ' 


2 


sin p sin(p a) 



These formulae are easily remembered on account of their 
similarity with the corresponding formulae of plane trigo- 
nometry. 

3. Relation between Two Sides and Their Opposite Angles. 3 
Apply the Law of Sines of plane trigonometry to the derived 
triangle relative to A (Fig. 11). 

a . b . c 
cos sm sm 



sin A 



sinE 



Used in Ex. 1, 18, 19, 20 (Ch. VIII). 
3 Used in Ex. 18, 21, 25, 27 (Ch. VIII). 



30 



IV. FOUR PARTS OF A SPHERICAL TRIANGLE 



Multiply both members by 2 sin - , 

i 

. a . b . c 

2 sin pr sm - sin s 

sin a _ 222 

sin A sin E 

In like manner, from the derived triangle relative to B, 
we get 

n . a . b . c 

. , 2 sin jr sin ^ sin ^ 
sin 6 __ 222 

sin B sin E * 

which was to be expected from the symmetrical form of the 
right-hand member of the above equations. Hence 



sm a 



__ sin b 
sin A ~~ sin B ' 



(9) 



This is expressed: The sines of the sides are proportional 
to the sines of the opposite angles. 

4. Relation between the Three Angles and One Side. 4 The 

Law of Cosines of plane trigonometry is applied to the triangle 
of elements of the polar triangle (Fig. 13). The method is 
the same as for the first formula (Sn. 1). 



cos A = cos B cos C + sin B sin C cos a. 



(10) 



This formula is easily remembered on account of its simi- 
larity with formula (7). Note the minus sign in the second 
member, however. 

5. Expression of the Half-Sides in Terms of the Three 
Angles. 6 The method used in Sn. 2 could be applied to the 
triangle of elements of the polar triangle (Fig. 13). Another 

Used in Ex. 27 (Ch. VIII). 
Used in Ex. 19 (Ch. VIII). 



OBLIQUE-ANGLED TRIANGLES 



31 



method does not require the use of the polar triangle. Draw 
the two key-triangles relative to B and those relative to C. 
The Law of Sines of plane trigonometry, applied to the 
triangle of elements relative to C, gives 



sin(A - E) 
sin C 



. a b 
sm cos 



sn 



the same law, applied to the derived triangle relative to B, 
gives 



sin E 
sin B 



. c . a 

Sin 2 sm 2 

b~ 

COS?: 



Multiplying these two relations by each other yields immedi- 
ately 



, a sin E sin(A E) 

i .. 

2 sin B sin C 



(lla) 



Likewise, the Law of Sines, applied to the triangles of 
elements relative to B and to C, yield two relations: one 
between B, C E, and the opposite sides; the other, between 
C, B E, and the opposite sides. These two relations, 
multiplied by each other, give 



cos 



sin(B - E) sin(C - E) 
sin B sin C 



(lib) 



Finally, the Law of Sines may be applied to both key- 
triangles relative to B, giving the relations 



sin E 



. c . a 
sm sm 



sin(B - E) c a ' 

cos cos ~ 



. a c 

/ A T-\ sin r cos o 
sm(A - E) 2 2 

sin(C - E) ~ . c a ' 
v 7 sm H cos H 



32 IV. FOUR PARTS OF A SPHERICAL TRIANGLE 

which, multiplied by each other, lead to 



2 a __ sin E sin(A E) 
2 ~ sin(B - E) sin(C - E) ' 



(He) 



The last formula could, of course, be derived from the pre- 
ceding two. 

The formulae (11) can only be remembered after careful 
comparison with the formulae (8). Note the deceiving simi- 
larity between (8a) and (lib), etc. 

6. Relation between Two Sides, their Included Angle, and 
the Angle Opposite One of them (that is to say, between four 
consecutive parts). 6 Napier's first two analogies, as read from 
the key-triangles relative to C, give 

. a - b 

f A-B Sin ~2~ .0 
tan ___ = __ n - cot _, 



sin 



tan 



A + B 



cos- 



2 

a b 



C 



cos 



A relation between a, 6, C, A can be obtained from these two 
equations by eliminating B, which is easily done as follows. 
Since 

A-B , A+B 



A - 

A. 



we may write 
tan A 



tan 



2 
A - B 



+ tan 



. 
1 tan 



A - 



^ 

tan 



2 ' 

A + B 

2 _ N 
A + B ~ D' 



Used in Ex. 2, 21, 27 (Ch. VIII). 



OBLIQUE-ANGLED TRIANGLES 



33 



where 



r . g-6 

sin _ 



N = 



cos 



a - b} 



sm 



cos 



a + b 



cot - 



o 

2 sm a cos ~- 

L 
_ 

sin(a + b) sin ^ 


sin a sin C 

Q 

sin (a + 6) sin 2 =- 



and 



sin(a - b) C 

D = 1 -- rf - r-TT COt 2 -pr 

sm(a + 6) 2 

C C 

sin(a + b) sin 2 -^ sin(a 6) cos 2 7^- 

2i 2t 

= _ 

sin (a + 6) sin 2 TT- 
J 

Substituting for N and D gives 

tan A 

__ sin a sin C 

a cos 6 ( sin 2 ~ cos2 o") +sin6cosaf sin 2 ^ 









sn 
whence 

sin 6 cos a sin a cos 6 cos C = sin a sin C cot A, 
or, on dividing by sin a and transposing, 



cot a sin b = cos 6 cos C + sin C cot A. 



(12) 



34 IV. FOUR PARTS OF A SPHERICAL TRIANGLE 

This is known as the "cot-sin-cos" formula. (Note the 
symmetry in the sequence of the trigonometric functions: 
cot-sin-cos . . . cos-sin-cot.) It is expressed: The cotangent 
of the side opposite one of the given angles times the sine of 
the other side is equal to the cosine of the latter times the 
cosine of the included angle, plus the sine of the latter times 
the cotangent of the angle opposite the first side. 

Remark. Formula (12) is sometimes written in another, 
just as elegant, form: 

cos b cos C = sin b cot a sin C cot A. (12a) 

It is then expressed: The product of the cosines of the side 
and the angle that are not opposite any given part is equal to 
the difference of their sines, each multiplied by a cotangent, 
respectively that of the given side and the given angle that 
are opposite each other. 

EXERCISES 

1. Express cosp in terms of the angles, by using Delambre's 
formulae and formula (7). Hint: 

P = Ja 4- J(& + c). 

2. Transform formula (lib) into formula (10). Hint: To elimi- 
nate E multiply both members by 2 and change the product of two 
sines in the numerator into a difference of two cosines. 

3. Derive formula (10) by applying formula (7) to the polar 
triangle. 

4. Derive the three formulae (11) by means of the triangle of 
elements of the polar triangle. 

5. Derive formula (12) from Napier's last two analogies (2). 

6. Write the formula giving cos c in terms of o, 6, C. Then replace 
cose in tne formula (7) by the value just found and, using the 
proportionality of the sines of sides and opposite angles, derive 
formula (12). 

7. Let 2S designate the sum of the angles of a spherical triangle. 
What form will the three formulae (11) assume with this notation? 



EXERCISES 35 

8. From formulae (8a) and (8b), find an expression of sin A. 
Show that it is equivalent to 



1 



sin A : Vl cos 2 a cos 2 6 cos 2 c + 2 cos a cos 6 cos c. 
sin 6 sin c 

*9. The expression of sin A, in terms of the sides, found from 
formulae (8a) and (8b) can also be obtained directly from the triangle 
of elements (Fig. 9), by applying to it the method used for deriving 
Cagnoli's formula (Ch. Ill, Sn. 5). 



CHAPTER V 

RIGHT-ANGLED TRIANGLES 

1. The Right-Angled Triangle, a Special Case of the Oblique- 
Angled Triangle. The formulae established in the preceding 
chapter give relations between four parts of a triangle. Of 
these parts, at least one is necessarily an angle. By letting 
an angle equal 90, a formula derived for the general case of 
an oblique-angled triangle is transformed into a relation be- 
tween three parts (other than the right angle) of a right-angled 
triangle. By convention l the triangle ABC is made right- 
angled at C. It may be necessary to rearrange the general 
formulae by permutation of letters, so that the angle which 
is to become the right angle be labeled C. The side opposite 
C will therefore be the hypotenuse c. 

There are six different ways of choosing three parts from 
the five parts (other than the right angle) of a right-angled 
triangle. The hypotenuse may be taken, either with the 
other two sides, or with the two angles other than the right 
angle, or with one side and one angle. In the latter case, the 
angle may be opposite the chosen side or adjacent to it. If 
the hypotenuse is not chosen, there are only two possibilities : 
two angles and one side, necessarily opposite one of the angles; 
or two sides and one angle, necessarily opposite one of 
the sides. 

Right-angled triangles are always with us. The six rela- 
tions about to be derived have therefore proved to be very 
useful in practice. You will find it advantageous to commit 
them to memory, preferably in the form of statements rather 
than equations. 

1 This convention is not universal. Many authors make A = 90. 

36 



PRODUCTS AND RATIOS 37 

2. Relations between the Hypotenuse and Two Sides or 
Two Angles. 2 The Product Formulae. 
Formula (7) may be written 

cos c = cos a cos 6 -f sin a sin b cos C. 
It becomes, 8 for C = 90, 



cos c = cos a cos b. 



(13) 



Formula (10) may be written 

cos C = cos A cos B + sin A sin B cos c. 
It becomes 4 



cos c = cot A cot B. 



(14) 



This is expressed: The cosine of the hypotenuse is equal 
to the product of the cosines of the two sides, or the product 
of the cotangents of the two angles. 

3. Relations between the Hypotenuse, One Side, and One 
Angle. The Ratios of Sines and Tangents. 
Formula (9) may be written 

sin c sin a 



sin C sin A " 
It becomes, 5 for C = 90, 

sin a = sin c sin A 

J The terms sides and angles, when applied to a right-angled tri- 
angle, are construed to mean sides other than the hypotenuse and 
angles other than the right angle. 

Formula (13) is used in Ex. 3, 7, 15, 17, 29 (Ch. VIII). 

* Formula (14) is used in Ex. 9, 12, 26 (Ch. VIII). 

5 Formula (15) is used in Ex. 4, 14, 23, 24, 29 (Ch. VIII). 



38 



V. RIGHT-ANGLED TRIANGLES 



or 



sin a 
sin c 


= sin A. 



(15) 



Formula (12) may be written 

cot c sin a = cos a cos B + sin B cot C. 
It becomes 6 



cos B = cot c tan a 



or 




(16) 



With respect to the side a, note that A is the opposite angle, 
whereas B is the adjacent angle. 

The relations are remembered as follows. In each case 
consider the side and the hypotenuse; the ratio of their sines 
is equal to the sine of the opposite angle, the ratio of their 
tangents is equal to the cosine of the adjacent angle. 

Remark. Compare the definitions of sine and cosine in a 
plane right-angled triangle: sin A' = cos B' = a'/c'. 

4. Relations between Two Angles and One Side, or be- 
tween Two Sides and One Angle. The Ratios of Cosines 
and Cotangents. 
Formula (10), 

cos A = cos B cos C + sin B sin C cos a, 
becomes, 7 on letting C = 90, 

cos A = sin B cos a 
or 



(17) 




Formula (16) is used in Ex. 4, 8, 20 (Ch. VIII). 
7 Formula (17) is used in Ex. 13 (Ch. VIII). 



NAPIER'S RULE 

Formula (12), 

cot a sin b = cos b cos C + sin C cot A, 
becomes, 8 for C = 90, 

cot a sin b = cot A 
or 




(18) 



The relations are remembered as follows. In each case 
consider an angle and its opposite side; the ratio of their 
cosines is equal to the sine of the other angle, the ratio of 
their cotangents is equal to the sine of the other side. 

Remark. Notice that all but one of the four ratios con- 
sidered in the last two sections give the sine of a part. 

5. Napier's Rule. Although some people prefer to memorize 
the preceding formulae as such, Napier's rule, which includes 
all the possible relations between any three parts of a right- 
angled triangle, may be found useful by others. 





FIG. 17. Napier's rule. 

In the triangle ABC, right-angled at C, ignore the right 
angle; then, the hypotenuse, the other two angles, and the 

8 Formula (18) is used in Ex. 7, 15, 17 (Ch. VIII). 



40 V. RIGHT-ANGLED TRIANGLES 

complements of the sides about the right angle are Napier's 
five cyclic parts. The rule is expressed as follows. 

The cosine of any of the five parts is equal to the product 
of the sines of the opposite parts or the product of the 
cotangents of the adjacent parts. 

cos c = cos a cos b = cot A cot B, 
cos A = sin B cos a = cot c tan b, 
cos B = sin A cos b = cot c tan a, 
sin a = sin c sin A = cot B tan &, 
sin 6 = sin c sin B = cot A tan a. 

This amazing rule is more than just a mnemonic trick. 
It is a theorem, which was given with a separate proof 9 
by Napier. 

It is easy to check that the ten relations yielded by the 
rule correspond, with some duplication, to the formulae (13) 
to (18). 

* Napier's proof is outside the scope of this book. 



CHAPTER VI 

EXAMPLES OF CALCULATIONS 

1. Solution of Triangles. 1 

(i) Oblique-angled triangles. 

Three parts being given, any fourth part may be calculated 
by one of the formulae (7) to (12). To find an angle in terms 
of the three sides, use one of the formulae (8), giving the half- 
angle. Likewise, to find a side in terms of the three angles, 
use one of the formulae (11), giving the half-side. In all 
other cases, use one of the fundamental formulae (7), (9), 
(10), (12). 

(ii) Right-angled triangles. 

Two parts being given, besides the right angle, any third 
part may be calculated by one of the formulae (13) to (18). 
Find the appropriate formula or use Napier's rule. 

An isosceles triangle is divided into two right-angled tri- 
angles by an arc drawn from the vertex at right angles to the 
base; this arc bisects the base and the opposite angle. Isos- 
celes triangles may also be solved like oblique-angled triangles 
by means of the fundamental formulae; formulae (8) and (11) 
should be avoided, as they would complicate the calculations. 

(iii) Right-sided (or quadrantal) triangles. 

The polar triangle of a right-sided triangle is right-angled; 
Napier's rule may be used to solve the polar triangle, from 
which the parts of the quadrantal triangle are then computed. 

1 We have seen (Introduction, Sn. 2) that any angle or any side of 
a spherical triangle is less than 180. It follows that a case of am- 
biguity will present itself whenever a part is to be calculated by a 
formula that gives its sine (two supplementary angles having 
equal sines). 

41 



42 VI. EXAMPLES OF CALCULATIONS 

Right-sided triangles are also easily solved by the funda- 
mental formulae, which are greatly simplified when one of the 
sides is equal to 90. 

2. Formulae Adapted to Logarithmic Computation. 

(i) FORMULA: cos a = cos b cos c + sin b sin c cos A. 
The unknown may be either a, b (or c), or A. If the un- 
known is a, factorize cos 6: 

cos a = cos 6(cos c -f tan 6 sin c cos A). 

Introduce an auxiliary angle u so as to arrive at the sine of 
the sum of two angles. 
Let 

cot u = tan b cos A, 
then 

cos b , . . . . 

cos a = (sm u cos c + sm c cos u) 

or 

cos 6 sin(c + u) 

cos a = r-^ ! - . 

sm u 

If the unknown is c, the same method gives immediately 

cos a sin u 



sin(c -f w) = 



cos b 



If the unknown is A, use the formulae (8), which are 
adapted to logarithmic computation. 

(ii) FORMULA: cot a sin 6 = cos b cos C -f- sin C cot A. 
If the unknown is a, we write 

A j. r / ~ , sin C cot A \ 

cot a = cot 61 cos C H r ) . 

\ cos 6 / 

Let 

. cot A 

cot v = r , 

cos 6 



LOGARITHMIC COMPUTATION 43 

then 

cot b , . ~ . . ~ x 

cot a = . - (sin v cos C + sin C cos v) 
sin t; 

or 

cot 6 sin(C + t;) 

- ~ - - 



. 
cot a = 



sin v 

If the unknown is C, the same method gives immediately 

cot a sin t; 



If the unknown is 6, we write 

A / r ,cosC\ . ~ , A 

cot al sin o cos b - 1 = sin C cot A. 
\ cot a / 

Let 

cos C 

tan w = 7 , 
cot a 

then 

COt d . 1 f \ SI ! A 

(sin o cos w sin w cos 6) = sm G cot A 

cos w 

or 

. ,, x sin C cot A cos w 

sm(6 w) = T . 

cot a 

If the unknown is A, the same method gives immediately 

, . cot a sin(6 w) 

cot A = : rr - . 

sin C cos w 

(iii) FORMULA: cos A = cos B cos C + sin B sin C cos a. 
The method is the same as for the first formula (Sn. 2, i). 
The results follow. 

Let 

cot x = tan B cos a. 

If the unknown is A, 

. cos B sin(C x) 

cos A = r-^ . 

sm x 



44 VI. EXAMPLES OF CALCULATIONS 

If the unknown is C, 

. ,~ x cos A sin x 

sm(C x) = 5 . 

cos B 

If the unknown is a, use the formulae (11), which are 
adapted to logarithmic computation. 

3. Numerical Calculations. 

Computations are carried out either by means of logarithms 
or with a calculating machine (in which case tables of natural 
values of the trigonometric functions are necessary). In 
either case, a neat calculation form is essential. 

(i) By logarithms. 

First write across the top of the page the formula to be 
employed (adapted to logarithmic computation, if desired). 
In this formula always designate the parts of the triangle by 
letters (not by their actual values). 

Then write the given parts near the left margin of the 
sheet. Avoid writing the word log as much as possible, use 
it only in front of those logarithms which may be needed 
again in the course of the calculations. Write the minus 
sign over a negative characteristic, 2 so as to avoid the cum- 

1 Calculations with negative figures are quite straightforward, once 
you become used to them. See for yourself: 

1.67 2.67 

2)3.34 2)3.34 

2__ _ 

13 13 

12 12 
14 14 

14 14 

In the division on the left, you say: 2 in 3 goes 1, 1X2 = 2, 
3 2 = 1, bring down 3, 2 in 13 etc. In the division on the 
right, you say: 2 in 3 goes 2, 2X2 = 1, 3 1 = 1, bring down 3, 
2 in 13 etc. The only difference is that, in the second case, you 
take the first partial quotient by excess in order to get a positive 
remainder. 



FIRST EXAMPLE 45 

bersome addition and subtraction of 10. In most cases there 
is no advantage in adding the cologarithm rather than sub- 
tracting the logarithm itself; in a simple division, especially, 
this amounts to making a subtraction and an addition instead 
of one subtraction only. 

Write the result near the left margin of the page. 

A second formula may be used as a check. This can be 
done in two ways: either the solution is carried out in dupli- 
cate, by means of two formulae; or a relation between the 
given parts and the result is verified afterwards. In the 
course of long calculations (involving a whole chain of tri- 
angles, for instance) it is well to check the results from time 
to time, before proceeding. 

(ii) With the calculating machine. 

Write the formula across the page. Copy only those 
natural values that may be used again in the course of the 
calculations. Write the data and the result near the left 
margin of the page. 

4. First Example. 

Given the three sides of a triangle, solve for one angle. 
Use a check formula. 



Again compare these two additions of logarithms : 

9.412 5062 -10 1.412 5062 

8.803 7253 -10 2.803 7253 



18.216 2315 -20 2.216 2315 

The last partial addition, in the example on the right, reads: 1 
(carried over) and I make 0, and 2 make 2. Is this really difficult? 



46 VI. EXAMPLES OF CALCULATIONS 

(A) SOLUTION BY LOGARITHMS 
. , A sin (p 6) sin (p c) , A sin p sin(p a) 

Sin 2 -pr : - : COS 3 7j- == -. j : . 

2 sin o sin c 2 sin o sm c 

a = 5048'20" 

b - 11644'50" (6315'10") T.950 8518 

c - 12911'40" (5048'20") 1.889 3049 

1.840 1567 
2p = 29644'50" 

171 

p - 14822'25 // (3137'35") T.719 6275 

p - a - 9734' 5" (8225'55") 1. 996 1989 

14 

T. 715 8449 
-T.840 1567 <- 
A T.875 6882 
log cos = 1.937 8441 
2 8462 
21 
12 1 
8 9 

- 2955'41.7" i 

Z 

p - 6 - 3137'35" .. T.719 6446 
p~c - 1910'45" .. T.516 5358 

303 
T.236 2107 

-T.840 1567 <- 

A T.396 0540 
log sin ~ = 1.698 0270 
2 0204 

66 
36 6 
29 4 

- 2955'41.8" 

6 

A - 5951'23.6" The answer to the closest second is 5951 / 24 // . 

Suggestion To find the supplement of an angle, write down 
the figures from left to right as you mentally increase the 



FIRST EXAMPLE 47 

given angle to 17959'(50/10)". Example: given 14822'25". 
Write 31 while saying 179, 37' while saying 59, 3 while 
saying 50, and 5" while saying 10. 

(B) SOLUTION BY NATURAL VALUES AND 
CALCULATING MACHINE 

. cos a cos 6 cos c 

COS A = - : r : - , 

sin o sin c 

cos sin 

a= 5048'20" .......... 0.6319542 .......... 

b 11644'50" .......... - 0.450 0551 .......... 0.893 0008 

c = 12911'40" .......... - 0.631 9542 .......... 0.775 0058 

cos A = 0.502 1668 
1817 
149 
125 7 

A 5951'23.6". 23 3 



i e i A * A sin(p 6) sin(p c) 
Check formula: tan 2 ^ = ~ - r ( 1 ~ ; . 
2 sin psm(p a) 

2p = 29644'50" 
p = 14822'25" (3137'35") .................... 0.524 3575 

p - a = 9734' 5" (8225 / 55 // ) ... 0.991 2859, . . . 206 

_ 32 0.524 3781 

0.991 2891 

p - b = 3137'35" .............................. 0.524 3781 

p - c 1910'45" .............. 0.328 5003 

_ 229 

0.328 5232 
tan 8 = 0.331 4101 



tan = 0.575 6824 
2 6708 
116 
64 6 
= 29 55'41.8" 51 4 

L 

A - 5951'23.6 /; . 



48 VI. EXAMPLES OF CALCULATIONS 

5. Second Example. 

Given the three angles of a triangle, solve for the three sides. 

(A) SOLUTION BY LOGARITHMS 

. . a sin E sin(A - E) . . b sin E sin(B - E) 

Sin 2 p: = - : ^5-^ ~ - - , Sin 2 - = - : ^ - - - '- 

2 sm B sm C 2 sm C sin A 

. . c sinEsin(C - E) 

Sin 2 - = - : - T : ^5 - . 

2 sin A sm B 

A = 6647' 0" ... T.963 3253 ... T. 963 3253 

B = 4230'40" . . T . 829 7752 ... T . 829 7752 

C = 9720'30". .T.996 4249 ... T. 996 4249 _ 

20638'10" T.826 2001 T.959 7502 T.793 1005 

2E - 2638'10" 
E = 1319' 5". .T.362 3558 . . . T.362 4003 . . . 1.362 4003 

445 

A - E = 5327'55"..1.904 9760 

78 
B - E - 2911'35" .............. T.688 1819 

188 
C - E = 84 1'25" . .......... ____ .......... . . T.997 6331 

T.267 3841 T.050 6010 T.360 0334 
-1.826 2001 -1.959 7502 -T.793 1005 
,.1.441 1840 h T.090 8508 ,1.566 9329 
log sin. ^T.720 5920 .5 T.545 4254 . T.783 4664 
2 5834 2 3938 2 4575 
86 316 89 
68 2 280 5 82 8 
17 8 35 5 62 



= 3142'12.5" 



a - 6324'25.0 /; 
b - 41 6'31.2" 
c - 74 48 ; 6.4" 



2p - 179 19' 2.6" 
Check formula : tan 2 -^ = tan ^ tan tan tan ^r 

i & & & & 



SECOND EXAMPLE 49 

p - 8939'31 . 3" ip - 4449'45 . 6" .... T . 997 3891 

211 
25 3 

p - a - 2615' 6.3" J(p - a) - 13 7'33.1" T.367 6676 

285 6 
9 5 

p - 6 - 4833 / 0.1" i(p - 6) - 2416'30.0" T.654 1690 
p-c= 1451'24.9" |(p-c) 725'42.4" T. 115 1788 

328 

65 6 

F 2.1344970 

F log tan ^ = 1.067 2485 

~ = 639'32.4" 2 2041 

444 
E = 1319' 4.8" 366 

78 

(fi) SOLUTION BY NATURAL VALUES AND 
CALCULATING MACHINE 

cos A + cos B cos C , cos B + cos C cos A 

COS a = : ^ : 7= . COS = : -^T-- 1 > 

sin B sin C sin C sm A 

cos C -f cos A cos B 

COS C = : 1 : r; . 

sm A sin B 

cos sin 

A 6647' 0" 0.394 2093 0.919 0207 

B = 4230'40" 0.737 1463 0.675 7332 

C = 9720'30" -0.127 7859 0.991 8018 

20638 / 10 // 

a = 6324'25.0" 0.447 6505 

6724 

219 

217 

2 

6 - 41 6'31.3" 0.753 4637 

4678 
41 

31 9 
9 1 

c 7448' 6.5" 0.262 1589 

1892 

303 
280 8 
22 2 



50 VI. EXAMPLES OF CALCULATIONS 

Check formula: 

E as * -f cos a -f- cos 5 4- cos c 

A a b c 
4 cos ~ cos o cos ~ 

0.850 7602 
177 8 

12 7 

Ja 3142'12.5" 0.850 7792 

0.936 3322 

68 

6 8 

i& 2033'15.6" 0.936 3397 
0.794 3852 
176 4 

23 5 

Jc 3724' 3.2" 0.794 4052 

cosE = 0.973 1061 

1119 

58 

56 

E = 1319 ; 5.2" 2 

2E 2638 / 10.4". 

6. Third Example. 

In a triangle ABC, right-angled at C, the following parts 
are known: a = 5936 / 30 // , b = 6422'. Find the angle A. 

(A) SOLUTION BY LOGARITHMS 

sin b = cot A tan a, 
hence 

tana 



tan A 



sin 6 



o - 5936 / 30" 0.231 7312 

6 6422 ; -T.955 0047 

log tan A - 0.276 7265 
167 

98 
A - 62 7'52". 



GEOGRAPHICAL PROBLEM 51 

(B) SOLUTION BY NATURAL VALUES 

. tana 1.705 0269 , QM 1COO 
tan A -T-T- = ^ nn* EQI^ "* 1 .891 1522 
sin 6 0.901 5810 



~428 
A = 62 7'52". 

7. Fourth Example. 

Given the geographical coordinates of two points A and B 
on a sphere, find the spherical distance between these two 
points. 

Let L and M, L' and M' be the longitudes and latitudes of 
the points A and B, respectively (Fig. 18). Let x be the 
arc AB. 




FIG. 18. Geographical problem. 

The formula (7), applied to the triangle NAB, immediately 
solves the problem. 

cos x = sin M sin M' + cos M cos M' cos(L L'). 
If the calculation is to be made by logarithms, write 
cos x = sin M'[sin M + cos M cot M' cos(L L')]. 

Let 

cot M' cos(L L') = tan w 

and compute the auxiliary angle w. 



52 VI. EXAMPLES OF CALCULATIONS 

We have then 

sin M' . _ .. , .,,. . s 

cos x = (sin M cos w + cos M sm w) 

cos w 

_ sin M' sin(M + w) 
cos w 

8. Numerical Application. 

Take the following geographical coordinates: New York, 
lat. 4043'N., long. 740'W.; San Francisco, lat. 3748'N., 
long. 12224'W. Assume the earth to be spherical and take 
its radius to be 3960 miles. 

What is the spherical distance (in degrees and minutes) 
between New York and San Francisco? What is the actual 
distance (in miles)? The coordinates being accurate to half 
a minute, with what precision can you give the distance? 

(A) SOLUTION BY LOGARITHMS 

Let L, L' be the longitudes of San Francisco and New York. 
Let M, M' be their latitudes. 

j.** f /T T f \ sin M'sin(M -f- w) 

tan w = cot M' cos(L L ), cos x = . 

v ' cos w 

M' 4043' 0.065 1775 

L 12224'30" 
L' 74 (T 0" 

L - L' 48 24'30" T.822 0487 

T.887 2262 
2024 
238 
217 5 
20 5 

w 3738'35.5" T.898 6243 

M 3748' 65 2 

M + w 7526 / 35.5 // 8 2 

log cos w = 1.898 6316 



NUMERICAL APPLICATION 53 

M' T. 814 4600 

M + ti; T.985 8270 

27 5 

2 8 

T.800 2900 

- 1.898 6316 

log coax = T.901 6584 

6649 

65 
x = 37 7'14.1" 636 

1 4 
The answer to the nearest half-minute is 37 7'. 

Length of the great circle on the earth (R = 3960 miles). 

2 0.301 0300 

TT 0.497 1499 

R 3.597 6952 

log27rR = 4.395 8751 
8678 

73 
70 

2wR = 24,881. 42 miles 3 

Length of 1, 1', 1" 

_ _ __ 

27rR 4.395 8751 4.395 8751 4.395 8751 

360 -2.556 3025 

21,600' -4.334 4538 

1,296,000" -6.112 6050 

1.8395726 0.0614213 2.2832701 

37 1.568 2017 

V 0.8450981 

14.1" 1.149 2191 

3.407 7743 0.906 5194 T.432 4892 

7647 82 83 

96 12 

37 - 2,557 . 25 The error on J' is more than i mile. 
7 8 . 0634 

14. 1" - 0.2707 The answer to the nearest half-mile 

2,565.58 is 2565* miles. 



54 VI. EXAMPLES OF CALCULATIONS 

(B) SOLUTION BY NATURAL VALUES AND 
CALCULATING MACHINE 

cos x = sin M sin M' + cos M cos M' cos(L L')- 

M 3748'.. 0.6129071 0.7901550 L = 12224'30" 

M' 4043' . . X 0.652 3189 X 0.757 9446 L' = 74 0' 0" 

X 0.663 8174 L - I/ = 4824'30" 

cosx = 0.797 3669 
3792 
123 
117 2 
3-37 7'14.2" 58 

Tables give the lengths of arcs to the radius 1. 

37 .... 0.645 7718 

7 ; .... 0.002 0362 J' .... 0.000 1454 

14 ;/ .... 0.000 0727 X 3960 

0.6478807 J' = 0.5758 miles 

X 3960 
x = 2565.6 

The answer to the nearest half-mile is 2565 i miles. 



CHAPTER VII 

PROBLEMS 

1. Let ABC be a tri-rectangular triangle (all angles and sides 
equal to 90) ; let p, q, r be the spherical distances from any 
point P inside the triangle to the vertices A, B, C, respectively 
(Fig. 19). Prove that 

cos 2 p + cos 2 q + cos 2 r = 1. 



90, 




FIG. 19. Distances from a point inside 
a tri-rectangular triangle to the vertices. 

In the triangle CAP, by formula (7), 
cos p = sin r cos x. 
Likewise, in the triangle BCP, 

cos q = sin r sin x. 
Squaring and adding, in order to eliminate x, we get 

cos 2 p + cos 2 q = sin 2 r = 1 cos 2 r, 
whence the desired relation. 

55 



56 VII. PROBLEMS 

2. A point P is located inside a tri-rectangular triangle ABC. 
Let x, y, z be its distances from the sides (Fig. 20). Prove 
that 

sin 2 x + sin 2 y + sin 2 2 = 1. 



90 




FIG. 20. Distances from a point inside 
a tri-rectangular triangle to the sides. 

Join PA, PB, PC and let these arcs be p, q, r. 

Let the arcs x, y, z intersect the sides BC, CA, AB in D, 
E, F. Let BD = u, CE = v, AF = w. 

The two right-angled triangles BDP and GDP yield the 
relations: 

cos q = cos x cos u t cos r cos x sin u. 
On squaring and adding, we get 

cos 2 x = cos 2 q + cos 2 r. 
Likewise, 

cos 2 y = cos 2 r + cos 2 p, 
cos 2 z = cos 2 p + cos 2 q. 

On addition, and in view of the fact that cos 2 p + cos 2 q + 
cos 2 r = 1, these three equations give 

3 - (sin 2 x + sin 2 y + sin 2 z) = 2, 
whence the desired relation. 



PROBLEMS 



57 



3. Let p, q, r and />', q' 9 r' be angles which two straight lines 
OP and OP' make with Cartesian coordinate axes (Fig. 21). 
Prove that the angle x between these two directions is 
given by 

cos x = cos p cos p' + cos q cos q' + cos r cos r'. 




FIG. 21. Angle between two directions (perspective drawing). 

Consider a sphere drawn with the origin as center. Let 
the radius of that sphere be taken as the unit of length. 
Produce the arcs ZP and ZP' to their intersections, Q and Q', 
with XY. Let XQ = y, XQ' = y'. 

In the triangle PZP', 



cos x cos r cos r' + sin r sin r' cos (y r y), 



or 



cos x = cos r cos r' + sin r sin r' cos y' cos y 

+ sin r sin r' sin ?/' sin y. 
In the triangle XPQ: 

cos p = sin r cos y. 



58 VII. PROBLEMS 

In the triangle XFQ': 

cos p f = sin r' cos y f . 
Likewise, in the triangles YPQ and YP'Q': 

cos q = sin r sin y, cos q' sin r' sin y'. 
Hence 

cos x = cos r cos r' + cos p cos p' + cos q cos #'. 

4. In a triangle ABC (Fig. 22), an arc x, drawn through one 
of the vertices C, determines on the opposite side c the arcs 
BD = p and AD = q. Prove that x is given by 

sin c cos x = sin p cos 6 + sin q cos a. 
c 




FIG. 22. Arc through one of 
the vertices. 

Let y designate the angle CDB. We have 

cos a = cos p cos x + sin p sin x cos y, 
cos b = cos <? cos x sin <? sin x cos y. 

Multiply the first equation by sin q, the second by sin p, 
and add the two together, so as to eliminate y. We get 

sin q cos a + sin p cos b = sin (p + q) cos x, 
which is the desired relation; and which may also be written 



PROBLEMS 59 

in the form 

sin p cos 6 + sin q cos a 

cos x = r^ - . 

sine 

Applications. 

(i) A median in terms of the sides. 

In this case, p q = ^c. The general formula becomes 

. c , . , x a + b a b 

sin jr (cos a + cos 6) cos ^ cos ~ 

<u L L 

COS X = = 

c 
sm c cos -= 

(ii) A bisector in terms of the sides. 

In this case, angle ACD = angle BCD = ^C. We have 

sin p __ sin a , sin q _ sin b 

sin C sin y sin C sin i/ ' 

whence 

sin p _ sin a 
sin <? sin 6 ' 

Now, since p + q = c, 

sin(c q) _ sin a 

sin q ~~ sin 6 ' 
which leads to 

, sin a + sin b cos c 

cot q = : r -. . 

sin 6 sm c 

Hence 

1 vsin b sin c 



sin q 



Vl + cot 2 q Vsin 2 a + sin 2 6 + 2 sin a sin 6 cos c 
Likewise (making q into p and interchanging a and 6), 

sin a sin c 



sin 



Vsin 2 a + sin 2 6 -f 2 sin a sin 6 cos c 



60 VII. PROBLEMS 

On substitution, the general formula becomes 

sin (a + b) 



cos x 



Vsin 2 a + sin 2 6 + 2 sin a sin b cos c 
(iii) An altitude in terms of the sides. 

In this case, y = 180 - y = 90. The general formula, 
thanks to the formulae of right-angled triangles 

cos a = cos x cos p and cos b = cos x cos q, 
can be simplified as follows 

cos a cos b 

cos x = = . 

cos p cos q 

From the relation 

cos p _ cos a 
cos q cos b ' 

by transformations similar to those used in (ii), we deduce 

, cos a cos b cos c 

tan q = j -7 , 

cos 6 sin c 

whence 



1 __ /. , , _ Vcos 2 a + cos 2 6 2 cos a cos 6 cos c 



^ J. | VlAlli U _ . 

cos q cos o sin c 

It follows that 



1 



cos x = - Vcos 2 a + cos 2 6 2 cos a cos b cos c. 
sin c 

5. A point P is located (Fig. 23) between two great circles 
AC and BC. The distances from P to these great circles are 
p and g, respectively. The angle ACB between the great 
circles is C. What is the distance x from P to the intersec- 
tion of the two given great circles? 

Let ACP = y, whence BCP = C - y. The two right- 
angled triangles give 

sin p sin x sin y, sin q = sin x sin (C y). 



PROBLEMS 

Eliminate y. The first equation gives 



sm p , 

sin y = -T - , whence cos y = 



Substituting in the second equation, we get 



61 



x """ 



sn 



sin q = sin C Vsin 2 x sin 2 p cos C sin p. 
c 




FIG. 23. Point between two 
great circles. 



Transposing and squaring, 



. , . , (sin q + cos C sin t>) 2 
sin' x - sm* p = ^^ , 



whence 
sin 2 x 



sin 2 p sin 2 C + sin 2 q + cos 2 C sin 2 p + 2 sin p sin g cos C 



sin 2 C 



and, finally, 



Sm X ~ s i n c Vsin 2 p + sin 2 5 + 2 sin p sin q cos C. 

Numerical Application. What is the latitude of a point 
in the Northern hemisphere, knowing its distances from the 
zero meridian (45) and from the 90-meridian (30)? 



62 VII. PROBLEMS 

Let L be the latitude. The above formula becomes 



cos L = Vsin 2 45 + sin 2 30, 

T ^ 
cos L = -- , 

whence 

L = 30. 

Remark. Note that, in this particular example, the point 
is located within a tri-rectangular triangle, so that (Problem 2) 

sin 2 45 + sin 2 30 + sin 2 L = 1. 

6. In a spherical triangle (Fig. 24), the cosine of the arc which 
connects the middle points of two sides is proportional to 
the cosine of half the third side ; the factor of proportionality 
is the cosine of one half of the spherical excess. 
Required to prove: 

*)' h 

cos x = cos E cos - , cos y = cos E cos ^ > 



cos z - cos E cos - . 




FIG. 24. Arc that connects the mid- 
points of two sides. 

Proof. In the triangle AEF, 

b c , . b . c , 
cos x = cos ~ cos 5 + sin ^ sm ^ cos A. 



PROBLEMS 63 

In the triangle ABC, 

cos a cos b cos c + sin b sin c cos A. 

Taking cos A from each equation, and equating the two 
values, we get 

b c 

COS X COS COS jr , 

2 2 cos a cos b cos c 



. b . c . . b . c b c' 

sm - sm x 4 sin ^ sin ^ cos ^ cos ~ 

L Jj A 



Hence 



cos a 
cosx = 



A C 

4 COS jr COS jr 



6 c 

cos a + 2 cos 2 o + 2 cos 2 



X 

4 COS pr COS ^ 

2i tli 

CQS a + (1 + cos 6) + (1 + cos c) 

^ 5 c 

4 cos - cos ^ 

z ^ 

1 -f cos a + cos 6 + cos c 

- T - . 

. 6 c 
4 cos ~ cos ^ 



This expression, divided by cos -= , is equal to cos E (Euler's 



formula) . Hence 



cos x = cos E cos ~ . 



Likewise for the other two formulae. 



64 VII. PROBLEMS 

*Second Proof. In a plane triangle 

c' = a' cos B' + b' cos A'. 

This formula, applied to the derived triangle (Fig. 11), where 
A' = 180 - A, B' = E, C' = A - E, gives 

be a ^ . b . c . 

cos 55 cos K = cos ~ cos E sm ^ sm ~ cos A 
2t & Z & 

or 

^ a b c , . b . c . 
cos E cos n = cos o cos 5 + sm - sm ^ cos A, 

A A i t 

which is equal to cos x, as seen in the triangle AEF (Fig. 24). 
Remark. The formula 

cos x = cos E cos ~ 

& 

can be derived directly in the following way. 




o 

FIG. 25. Geometrical construction 
(perspective drawing). 

Construction. Join (Fig. 25) the vertices A, B, and C to the 
center of the sphere of unit radius (OA = OB = OC = 1). 



PROBLEMS 65 

Join to the middle points E and F of the sides b and c. 
Draw the chords AB, BC, and CA. Join H, the point of 
intersection of OE and CA, to K, the point of intersection 
of OF and AB. 

Proof. The formula giving a relation between the three 
sides and one of the angles of a plane triangle is applied to 
the triangle OHK. 

Since the angle COA = 6, we have, in the plane OCA, 



Likewise 



OH = cos ~ . 



/ 
OK = COS j: . 



Since E is the middle point of the arc CA, H is the middle 
point of the chord CA. Likewise, K is the middle point of 

C*H\ n. 

AB. Hence, in the triangle ABC, HK = ^ = sin ~ , for 

A & 

the chord subtending an arc is equal to twice the sine of half 



the arc ( CB = 2 sin | j . 



Now the angle EOF is measured by the arc EF = x. 
By plane trigonometry, we have, in the triangle OHK, 

sin 2 jr = cos 2 ~ + cos 2 ~ 2 cos ^ cos ~ cos #, 

A U Li & 4 

whence 

2 cos 2 ~ + 2 cos 2 | - 2 sin 2 ~ 

cos x = z , 

A b c 
4 cos cos ^ 

& A 

(1 + cos 6) + (1 + cos c) (1 cos a) a 

cos x = -^ ^- i -^ L T cos x , 

. a b c a 2 ' 

4 COS <r COS ^ COS ^ COS ^ 
COS X = COS E COS x . 



66 VII. PROBLEMS 

7. The area of an equilateral triangle is S/n, where S desig- 
nates the area of the sphere. Find the side x of the triangle. 

Let A be the angle of the triangle. By Girard's theorem 
(Appendix I), 

1 }(3A - 180) , . n + 4 - no 

_ - * L whence A = 60 . 

n 360 * n 

The side x is given in terms of the angles by the fundamental 
formula (10) 

cos A = cos 2 A + sin 2 A cos x, 

from which we deduce 

cos A cos A 



cos x 



1 cos A . 9 A 
2 sm* 



or, after substitution, 



COS X = 



Numerical Application. Take n = 4. Show that x *arc 
cos ( -J). 

cos 120 -\ , 



x = 180 y, where cos y = J. 

log cosy = 1.522 8787 
9002 
215 
178 5 
36 5 

y = 7031'43.6" 
x = 109 28'16.4". 



PROBLEMS 67 

8. Given the dihedral angle (or edge angle) a between two 
slant faces of a regular n-sided pyramid, calculate the face 
angle x between two adjacent slant edges (Fig. 26). 




s 
Fio. 26. A regular pyramid. 

The base of the pyramid is an n-sided regular polygon; 
the interior angle between two sides is equal to 

y = ^-? 180. 

n 

Around the vertex S draw the unit sphere, so as to replace 
the trihedron S by a spherical triangle. Decomposing this 
isosceles triangle in two right-angled triangles, we get im- 
mediately 



whence 



. y . ot x 
sm ^ = sm ^ cos ~ > 



sin Q()0 

x n 

COS^r 



2 . a 

sm- 



68 VII. PROBLEMS 



x 
Discussion. The problem is possible only if < cos ~ < 1- 

4U 

X X 

It is obvious that cos cannot be negative, as must be 



) 



smaller than 90. ) The first condition demands 

sin 90 > 0, 

n ' 

which implies n >: 3. ( Note that sin -= is always positive, j 
The second conditions requires 

sin 75 > sin 90, 

A n 

that is to say a > 180, which is evident (by geometry). 

Numerical Application. In a regular hexagonal pyramid, 
let a = 170. The general formula becomes 

x sin 60 1 .937 5306 

COS 2 sin 85 1.998 3442 

log cos I = 1.939 1864 

1953 

89 

84 

5 

I = 2937'7.4" 
x = 5914'14.8". 

9. Volume of a parallelepiped. 

Let OA = a, OB = 6, OC = c, be three edges making with 
one another (Fig. 27) the angles BOG = a, COA = ft 
AOB = 7. Through C pass a plane normal to OA and con- 
taining CD, perpendicular to the plane AOB. The angle 



PROBLEMS 69 

CED = A is the dihedral angle of the edge OA. Draw the 
unit sphere around the vertex 0. 




FIG. 27. Volume of a parallelepiped. 

V = ab sin y . CD = ab sin 7 . CE sin A = abc sin 7 . sin /3 sin A. 
But 

A A 

sin A = 2 sin cos 



= 2 Vsin p sin (p a) sin (p ff) sin (p 7) 
sin /3 sin 7 ' 

in which 2p = a + + 7, from the formulae (8). 
It follows that 

V = 2a6c Vsin p sin (p a) sin (p 0) sin (p 7). 

Special Cases. 

Volume of a rhombohedron (b = c = a, /? = 7 = a) : 



TT- r 7 

V = 2a 3 



. 

sin 3 



a 
-= . 



Volume of a right-angled parallelepiped (a = = 7 = 90) : 

V = a6c. 

Volume of a cube (6 = c = a, a = j3 = 7 = 90) : 

V = a 3 . 



CHAPTER VIII 

EXERCISES 1 

1. The face angles of a trihedron measure 50. Calculate one of 
the dihedral angles (edge angles) . 

2. In a triangle ABC, given b = 45, c = 60, A = 30. Find B. 

3. Two triangles ABC and BCD, both right-angled at C, have a 
side of the right angle in common, BC = 5936'30". The other 
side about the right angle is known in each case: CA = 6422', 
CD 52 5'. Find the hypotenuses AB and BD. 

4. Given, in a right-angled triangle (C = 90), cose = 1/V3 and 
b = 45. Calculate the other two angles, each from the data only, 
and express the area of the triangle as a fraction of the area S of the 
sphere on which it is drawn. 

5. Show that if (C - E) : A : (B - E) = 2 : 3 : 4, then 

2 cos i(C - E) sin %(b + c)/sin \a. 

[Hint: it is known that if the angles of a plane triangle A'B'C' are 
in the ratios A' : B' : C' = 2 : 3 : 4, then 2 cos JA' = (a' + c')/b'.] 

6. An equilateral triangle ABC, with a = b = c = 30, is drawn 
on a sphere of radius R. Find the area of the triangle expressed in 
dm 2 if R = 6w. 

7. Let O, O', O" be the middle points of the three edges of a cube 
intersecting in the same vertex C. Calculate the elements of the 
trihedron whose edges are OC, 00', OO". 

8. In an isosceles triangle, given b = c = 32 and B = 78. Find 
the side a. 

9. In a regular tetragonal pyramid, the altitude is equal to one 
half the diagonal of the base. Calculate the dihedral angle between 
two adjacent slant faces. 

10. Express sin (a 4- b) /sin c in terms of the three angles of an 
oblique-angled triangle. 

1 Most of these exercises are questions that were asked by Cesaro 
at the entrance examination of the University of Lie"ge. 

70 



EXERCISES 71 

11. The arc x that connects the middle points of two sides of an 
equilateral triangle of side a is given by 2 sin \x tan Ja. (Compare 
Problem 6, Chapter VII.) 

12. In a regular tetragonal pyramid, the angle of a face at the 
apex measures 60. Calculate the dihedral angle between two ad- 
jacent slant faces. 

13. In a regular pentagonal pyramid, the angle of a face at the 
apex measures 30. Calculate the dihedral angle between two ad- 
jacent slant faces. 

14. The altitude CH of a triangle ABC, right-angled at C, inter- 
sects the hypotenuse AB at H; arc AH = b' and arc BH = a'. 
Prove that 

*t "' sin 2 b' _ 

I . n T ~ 1 . 



sin 2 a sin 2 6 

15. Given a tri-rectangular trihedron O,XYZ. Plot OP = 2, 
OQ = 3, OR = 4, on the edges OX, OY, OZ, respectively. Calcu- 
late the elements of the trihedron P, OQR. 

16. Let ipi> PI be the longitude and polar distance (or colatitude) 
of a point PI; <?A, PA, those of a point A. A 180-rotation about A 
brings PI in P2. What are the longitude <pt and the polar distance 
P2 of P 2 ? 

17. Let AB - p, AC = q, AD = r be three edges of a right- 
angled parallelepiped . Calculate the elements of the trihedron whose 
edges are DA, DB, DC. Numerical application: p = 3, q = 4, 
r = 5. 

18. In a quadrantal triangle (c = 90), given C and 2p. Find 
(1) sin a sin 6; (2) sin A sin B. 

19. In a right-angled triangle (C = 90), given the hypotenuse c 
and the spherical excess 2E. Find: (1) sin a sin 6; (2) sin A sin B. 

20. Given an equilateral triangle whose side is equal to 60. Find 
the side of the triangle formed by erecting, at each vertex, a perpen- 
dicular to the corresponding bisector. 

21. Solve a quadrantal triangle (a = 90), given: (1) c, C; (2) 
A, b + c. 

22. Prove the formulae of a quadrantal triangle (a = 90) : 

sin p SB cos JB cos iC/sin iA, 
cos p = sin JB sin iC/sin JA, 
tan p = cot iB cot JC. 



72 VIII. EXERCISES 

23. Consider a regular pentagonal prism with altitude h and the 
side of the base equal to a. Join 0, one of the base vertices, to P 
and Q, the top vertices opposite the two adjacent vertical faces. 
Calculate the angle POQ. Numerical application: a = 4, h = 8. 

24. Same question for a hexagonal prism. Numerical application : 
a = 5, h - 5. 

25. Join a point P, inside an oblique-angled triangle ABC, to the 
vertices and produce the arcs AP, BP, CP to their intersections, D, 
E, F, with the sides of ABC. Prove that 

sinCE sin AF sin BD 
sin EA sin FB sin DC 

26. In a tri-rectangular triangle ABC, join the middle points B' 
and C' of the sides AB and AC by the arc B'C'. Find the ratio of 
the areas of the triangles AB'C' and ABC. 

27. In a triangle ABC, let a, 0, 7 be the lengths of the bisectors. 
Prove: 

(1) cot a cos JA -f cot /3 cos JB -f- cot 7 cos JC 

= cot a -f- cot b -f- cot c; 
cot a cot j8 



(2) 



cos JA(cos B H- cos C) cos JB(cos C -f- cos A) 

cot 7 



cos iC(cos A -f cos B) ' 



. 2 Vsin b sin c sin p sin (p a) 



Vsin 2 6 + sin 2 c -f- 2 sin 6 sin c cos a 

28. In a triangle ABC, given a - b 90 and C 60. Join any 
point P inside the triangle to the vertices A, B, C by arcs x, y, z. 
Find a relation between x, y y z, 

29. In a quadrilateral ABCD, given :B C = D = 90, BD = a, 
and AB b. Calculate the diagonal AC. 

30. Show that the radii R c , R, Ra of the circumscribing, inscribed, 
and escribed circles of a spherical triangle are given respectively by: 

2 P - cosS 

tan KC 



where 2S = A + B -f C; 



EXERCISES 73 

sm(p a) sin(p 6) sin(p c) 

where 2p a 4- b + c; 



tan 2 R % 

sin p 



tan 2 R = sin p sin(p - 6) sin(p c) 
sin(p a) ' 

where R is the radius of the escribed circle which touches a. 



ANSWERS TO EXERCISES 

Numerical applications have been calculated with five-place tables. 
The answer is followed, between parentheses, by the number of the 
formula to be applied. 

1. 6658' (8). 

2. B = 496'24" (12). 

3. AB = 7721'28", BD = 7153'14" (13). 

4. A = 60, B = 45, T = S/48 (15 and 16, no tables of logarithms 
needed). 

5. The desired relation is read off the triangle of elements (Fig. 9). 

6. E = 3.5, T = 439.8 dm 2 (5a and Appendix 1). 

7. COO' = COO" = 45, O'OO" = 60, edge angle OC = 90, 
edge angle OO' = edge angle OO" = 5444' (13 and 18). 

8. a = 1448'17" (16). 

9. 10928' (14). 

10. sin(a -f b)/sin c = (cos A -f cos B)/(l - cos C). Apply (3) (4). 

11. Apply (5a). 

12. 10928' (14). 

13. 11346' (17). 

14. Apply (15) twice. Use sin z z + cos 2 a; = 1. 

15. Edge angle PO = 90, OPR = arctan 2 = 6326' 5", OPQ 
- arctan (3/2) = 5618'35", edge angle PR = 59 11 '33", edge angle 
PQ = 6724 / 41", RPQ - 7538'11". Apply (13) and (18), the 
latter twice. 

16. cos p2 = cos 2pA cos pi + sin 2pA sin pi cos (V?A ^0, 
sin (<f>2 <PA) sin pi sin (V>A ^0/sin p 2 . Apply Problem 4, Ch. 
VII and (7). (This question is encountered in crystallography.) 

17. The spherical triangle ABC is right-angled at A. Its parts 
are: a = 4757 / 51 // , b = 3839 r 36", c = 3057'50 // , B = 5715 / 15", 
C 4350'17 /; . Apply (13) and (18). First, determine 6 and c by 
plane trigonometry. 

18. sin a sin b = sin 2p/(l + cos C), sin A sin B = sin 2p 
(1 cos C). Apply (8b), then combine with (9). 

74 



ANSWERS TO EXERCISES 75 

19. sin a sin b = sin2E(l + cos c), sin A sin B = sin2E/(l cose). 
Apply (11) and (8), or consider polar triangle of quadrantal triangle 

in preceding exercise. 

20. 90 (8a, then 16; no tables of logarithms needed). 

21. Given c, C. Write sin & = cosc/cosC, sin A = sinC/sinc, 
sin B = cot c/cot C. Apply (7) (9) (12). 

Given A, b + c. Express cos i(B C) and cos i(B -f C) in 
terms of J(& -f c) and JA by means of (3) (4), hence B and C. Like- 
wise express sin i(& c) in terms of $(B C) and JA by means of 
(3) or (4). Since (6 -\- c) is known, you get 6 and c. 

22. Express the corresponding relations in the (right-angled) polar 
triangle of the given triangle. Apply the Law of Sines to Fig. 11. 

23. sin i(POQ) = sin 18 sm c, with tan c 2(a/h) cos 36. Ap- 
ply (15). For a = 4 and h = 8, POQ = 2224'53". 

24. Apply (7) or (15). For a/h = 1, POQ = 5119' 4". 

25. Apply (9). Note three pairs of equal angles having common 
vertex P. (This question is encountered in crystallography.) 

26. 0.216 (14 or 7, and Girard's Theorem, Appendix 1). 

27. (1) Apply (12). (2) Apply (10) (9). (3) Apply (12) twice. 

28. 4 (cos 2 x -f cos 2 y cos x cos y) = 3 sin 2 z. Apply (7) and 
sin 2 a -f cos 2 a = 1. 

29. Let x = AC. It is given by the biquadratic equation cos 4 x 
cos 2 a cos 2 b cos 2 x cos 2 a cos 2 b sin 2 6 = 0. Apply (13) three 
times and (15) twice. 

30. Consider a circle inscribed in a spherical triangle. It is tangent 
to the sides, The spherical distances from a vertex to the nearest 
two points of tangency are equal. The radii to the points of tangency 
are perpendicular to the corresponding sides. Apply the formulae 
of right-angled triangles. 



APPENDIX 1 
SPHERICAL AREAS 

1. Sphere. The area S of a sphere is equal to four times that of a 
great circle, that is to say S = 4irR 2 , if R designates the radius. 

2. Lune. The area of a lune of angle A is to the area of the sphere 
as A is to 360. 

Consider a number of meridians; they intersect in the North and 
South Poles. Any two meridians bound a lune (for example, the 
meridians 75W and 80W). The angle of the lune is the angle 
between the two meridians (in this case, 5); it is measured by the 
arc they intercept on the equator. Obviously lunes of equal angles 
have equal areas (for they can be made to coincide). The areas of 
lunes are proportional to their angles (this is proved in the same 
manner as the proportionality of angles at the center of a circle to 
the arcs they subtend), whence the desired proposition. 

3. Spherical Triangle. GIRARD'S THEOREM: The area of a spher- 
ical triangle is to the area of the sphere as half the spherical excess 
(expressed in degrees) is to 360. 




FIG. 28. Area of a spherical triangle. 

Consider a triangle ABC. Join its vertices A, B, C to the center O 
of the sphere (Fig. 28) and produce to the antipodes A', B' f C'. 
Complete the great circles that form the sides of the triangle ABC. 

76 



SPHERICAL AREAS 77 

The hemisphere in front of the great circle BCB'C' is thus divided 
into four triangles: 

Ti ABC, T 2 AB'C', T, - AB'C, T 4 = ABC'. 

From the construction it follows that the triangle A'BC is equal to 
the triangle T 2 =* AB'C'. Hence, designating by lune A the area 
of a lune of angle A, 

Ti -f T 2 = lune A. 
Moreover 

Ti + T, = lune B 
and 

Ti + T 4 = lune C. 
On adding and transposing, 

2 Ti = lune A -f lune B + lune C - (Ti + T 2 -f T 3 -f T 4 ), 
whence 

2Ti A + B + C - 180 ^ 2JS 
8 360 " 360 

and, finally, 

II JL 
S 360* 



APPENDIX 2 
FORMULAE OF PLANE TRIGONOMETRY 

DEFINITIONS: versine, coversine, exsecant, coexsecant, haversine. 

vers x = 1 cos x, exsec x = sec x 1 , 

co verso: = 1 sin x, coexsecx = cscx 1, 

hav x = J vers x = i(l cos x) = sin 2 Jx. 

sin 30 = i, sin 45 = ^ , sin 60 = ^? , sin 90 = 1. 

/L i 

Complementary angles, x and (90 x), have complementary 
trigonometric functions, with the same sign: 
sin x = cos (90 - x), tan x = cot (90 - x), sec x = esc (90 x). 

Supplementary angles, x and (180 x), have the same trigo- 
nometric functions, with opposite signs except sine and cosecant. 

Antisupplementary angles, x and (180 -f x), have the same trigo- 
nometric functions, with opposite signs except tangent and cotangent. 

Equal angles of opposite signs, x and x, have the same trigo- 
nometric functions, with opposite signs except cosine and secant. 



1 cot x 

cosx = 



Vl + cot 2 x ' 

sin (x y) = sin x cos y db sin y cos x, 
cos (x db y) = cos x cos y =F sin x sin #. 

, N tan x . tan y 



sin 2x = 2 sin x cos x, 2 sin 2 Jx = 1 cos x, 
cos 2x = cos 2 x sin 2 x, 2 cos 2 Jx = 1 + cos x, 

, 2 tan x x o i 1 cos x 

tan 2x = ; - r r , tan 2 \x = ^ ; - . 
1 - tan 2 x ' 3 1 4- cos a? 

78 



FORMULAE OF PLANE TRIGONOMETRY 79 

1 -f- tan x ,AKO , \ cot as -t- 1 , //4 _ x 

- IT = tan (45 H- x), ? = cot (43 x). 

1 tan a; v " cot x 1 v ' 

2 sin x cos y = sin (x -f- y) -f- sin (x y), 
2 cos x cos 2/ cos (x -f- ?/) -f- cos (x y}, 
2 sin x sin ?/ = cos (x y) cos (x + 2/) 

sin P + sin Q = 2 sin J(P -f- Q) cos i(P - Q), 
sin P - sin Q = 2 cos ^(P + Q) sin J(P - Q), 
cos P -f cos Q = 2 cos i(P -f Q) cos J(P - Q), 
cos P - cos Q = 2 sin $(P + Q) sin KQ - P). 

cos P cos Q . . /T ^ , ,^.1. . , /rA -TJV 
- tan i(P + Q) tan i(Q - P) ' 



SOLUTION OF TRIANGLES 
a b c 



LAW OF SINES: 



sin A sin B sin C " 
c 6 cos A 



Corollary: cot V = 

LAW OP COSINES: o 2 = b 2 + c 2 26c cos A. 

T _, tan KB - C) tanKB-C) b - c 

LAW OP TANGENTS: tan >(B + c) - cot iA = 5^^ 



Half -angle formulae: - 
(2p = a -f- 6 + c) 



cos 2 - = p(p r "" a) , 
2 be 

A (p-6)(p -c) 
TT = - 7 - r - . 
2 p(p a) 

Area o/^e triangle: T = Vp(p a)(p 6)(p c). 



INDEX 



Altitude 60 

Ambiguity 41 

Analogies 22 

Angle between two direc- 
tions 57 

Angular distance 5, 51 

Antiparallel 8 

Antisupplementary angles . 78 
Arc through one of the ver- 
tices 58 

Area of a lune 76 

Area of a sphere 76 

Area of a spherical triangle 76 
Astronomy 6 

B 

Bisector 59 



Cagnoli's formula 25 

Calculations 41, 44 

Cesaro 70 

Cesaro's key-triangles 13 

Checking calculations 45 

Circumscribing circle 72 

Coexsecant 78 

Complementary angles. ... 78 

Concyclic points 15 

Consecutive parts 32 

"Cot-sin-cos" formula 34 

Coversine 78 

Crystallography 5, 74, 75 



Cube 69 

Cyclic parts 40 



Delambre's formulae 22 

"Derived triangle" 16 

Dihedral angle 3 



Equal angles of opposite 

signs 78 

Escribed circle 72 

"Eulerian" spherical tri- 
angle 4 

Euler's formula 23 

Examples of calculations . . 45 
Exsecant 78 

F 

Fundamental formulae. ... 41 



Gauss 22 

Geodesy 5 

Geographical problem 51 

"Geoid" 6 

Geometry 5 

Girard's theorem 66, 76 

Great circle 3 

H 

Half-angles 29 

Half-sides 30 

Haversine 78 



81 



82 



INDEX 



Hypotenuse 36 

Hypotenuse and two sides 

or two angles 37 

Hypotenuse, one side, and 

one angle 37 



Inscribed circle 72 



Point between two great 

circles 61 

Polar triangles 18 

Poles 18 

Polygon 3 

Precision 52 

Problems (with solutions) . . 55 

Product formulae 37 

Pyramid 67 



Law of cosines 79 

Law of sines 79 

Law of tangents 79 

Lhuilier's formula 24 

Logarithmic computation, 

formulae adapted to 42 

Lune 16, 76 

M 

Median 59 

Mid-points of two sides ... 62 

N 

Napier's analogies 22 

Napier's rule 39 

Navigation 6 

Negative characteristic 44 

O 

Oblique-angled spherical 
triangles 28, 41 



Parallelepiped 68 

Perimeter 4 

Plane triangle 20 

Plane trigonometry formu- 
lae 78 



Quadrantal triangles 41 



Ratios of cosines and co- 
tangents 38 

Ratios of sines and tangents 37 

Rhombohedron 69 

Right-angled parallelepiped 69 

Right-angled triangles. 36, 41 

Right-sided triangles 41 



Sine of the trihedral angle . 21 

Solution of triangles 41 

Spherical areas 76 

Spherical distance 5, 51 

Spherical excess 14, 23, 

24, 25, 26 

Spherical polygon 3 

Spherical triangle 4, 21 

Stereographic projection . . 7 

Sub-contrary 8 

Supplementary angles 78 

Surveying 5 



Three angles and one side . 30 
Three sides and one angle . 28 



INDEX 83 

"Triangle of elements". ... 13 Two sides and opposite 

Trihedra, complementary . 16 angles 29 

Trihedron 4 Two sides, their included 

Tri -rectangular triangle. . 55, angle, and the angle op- 

56, 62 posite one of them 32 

Two angles and one side, or 

two sides and one angle 

(in a right-angled tri- Versine 78 

angle) 38